buffers solutions

5/23/2018 Buffers Solutions

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Titration curves

A titrationis a procedure for carrying out a chemical reaction between two solutions by thecontrolled addition from a buret of one solution (the titrant) to the other, allowing

measurements to be made throughout the reaction. For a reaction between an acid and a base,

a titration is useful for measuring the pH at various points throughout the reaction.

A titration curveis a graph of the pH as a function of the amount of titrant (acid or base)

added.

Strong Acid-Strong Base Titrations

Here is an example of a titration curve, produced when a strong base is added to a strong acid.

This curve shows how pH varies as 0.100 M NaOH is added to 50.0 mL of 0.100 M HCl.

The equivalence pointof the titration is the point at which exactly enough titrant has been

added to react with all of the substance being titrated with no titrant left over. In other words,

at the equivalence point, the number of moles of titrant added so far corresponds exactly to

the number of moles of substance being titrated according to the reaction stoichiometry. (In an

acid-base titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point

where the moles of titrant added equals the moles of substance initially in the solution being

titrated.)

Notice that the pH increases slowly at first, then rapidly as it nears the equivalence point.

Why?


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Learning Goal 25

Calculate the pH at any point, including the equivalence point, in an acid-base titration.

At the equivalence point, the pH = 7.00 for strong acid-strong base titrations. However, in

other types of titrations, this is not the case.

EXAMPLE:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of

0.100 M HCl solution?

Because it is a strong acid-base reaction, the reaction simplifies to:

H+(aq) + OH

-(aq) H2O (l)

The original number of moles of H+in the solution is:

50.00 x 10-3

L x 0.100 M HCl = 5.00 x 10-3

moles

The number of moles of OH-added is :

49.00 x 10-3

L x 0.100 M OH-= 4.90 x 10

-3moles

Thus there remains:

(5.00 x 10-3

) - (4.90 x 10-3

) = 1.0 x 10-4

moles H+

(aq)

The total volume of solution is 0.04900 L + 0.05000 L = 0.09900 L

[H+] = {1.0 x 10

-4moles / 0.09900 L } = 1.0 x 10

-3M

pH = 3.00

Titrations Involving a Weak Acid or Weak Base

Titration curve of a weak acid being titrated by a strong base:


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Here, 0.100 M NaOH is being added to 50.0 mL of 0.100 M acetic acid.

There are three major differences between this curve (in blue) and the one we saw before (in

black):

1. The weak-acid solution has a higher initial pH.

2. The pH rises more rapidly at the start, but less rapidly near the equivalence point.

3. The pH at the equivalence point does not equal 7.00.

POINT OF EMPHASIS :The equivalence point for a weak acid-strong base titration has a

pH > 7.00.

For a strong acid-weak base or weak acid-strong base titration, the pH will change rapidly at

the very beginning and then have a gradual slope until near the equivalence point. The gradual

slope results from a buffer solution being produced by the addition of the strong acid or base,

which resists rapid change in pH until the added acid or base exceeds the buffer's capacity and

the rapid pH change occurs near the equivalence point.

EXAMPLE:

What is the pH when 30.0 mL of 0.100 M NaOH have been added to 50.0 mL of 0.100 M

acetic acid?

STEP 1:Stochiometric calculation:

The original number of moles of HC2H3O2in the solution is :

50.0 x 10-3

L x 0.100 M = 5.00 x 10-3

moles HC2H3O2


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Similarly, there are 3.00 x 10-3

moles of OH-due to the NaOH solution.

The reaction goes to completion:

OH-(aq) + HC2H3O2(aq) C2H3O2

-(aq) + H2O (l)

OH- HC2H3O2 C2H3O2-

INITIAL 3.00 x 10-3

mol 5.00 x 10-3

mol 0

CHANGE -3.00 x 10-3

mol -3.00 x 10-3

mol +3.00 x 10-3

mol

FINAL 0 2.00 x 10-3

mol 3.00 x 10-3

mol

The total volume is 80.0 mL.

We now calculate the resulting molarities :

[HC2H3O2] = { 2.00 x 10-3mol HC2H3O2/ 0.0800 L } = 0.0250 M[C2H3O2

-] = { 3.00 x 10

-3mol C2H3O2

-} / 0.0800 L } = 0.0375 M

STEP 2:Equilibrium calculation, using simplification:

Ka= { [H+][C2H3O2

-] / [HC2H3O2] } = 1.8 x 10

-5

[H+] = { KA[HC2H3O2] / [C2H3O2

-] } = { (1.8 x 10

-5)(0.0250) / (0.0375) } = 1.2 x 10

-5M

pH = -log(1.2 x 10-5

) = 4.92

Titration curve of a weak base being titrated by a strong acid:

Here, 0.100 M HCl is being added to 50.0 mL of 0.100 M ammonia solution.


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As in the weak acid-strong base titration, there are three major differences between this curve

(in blue) and a strong base-strong acid one (in black): (Note that the strong base-strong acid

titration curve is identical to the strong acid-strong base titration, but flipped vertically.)

1. The weak-acid solution has a lower initial pH.

2. The pH drops more rapidly at the start, but less rapidly near the equivalence point.

3. The pH at the equivalence point does not equal 7.00.

POINT OF EMPHASIS :The equivalence point for a weak base-strong acid titration has a

pH < 7.00.

Titrations of Polyprotic Acids

An example of a polyprotic acid is H2CO3which neutralizes in two steps:

H2CO3(aq) + OH-(aq) H2O (l) + HCO3

-(aq)

HCO3-(aq) + OH

-(aq) H2O (l) + CO3

2-(aq)

The titration curve for these reactions will look like this, with two equivalence points.

Uses of Titrations

Learning Goal 26

Use titration data or a titration curve to calculate reaction quantities such as the concentration


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of the substance being titrated.

The most common use of titrations is for measuring unknown concentrations. This is done by

titrating a known volume of the unknown solution with a solution of known concentration

(where the two react in a predictable manner) and finding the volume of titrant needed to

reach the equivalence point using some method appropriate to the particular reaction. Then,

the volume and concentration of titrant can be used to calculate the moles of titrant added,

which, when used with the reaction stoichiometry, gives the number of moles of substance

being titrated. Finally, this quantity, along with the volume of substance being titrated, gives

the unknown concentration.

For acid-base titrations, the equivalence point can be found very easily. A pH meter is simply

placed in the solution being titrated and the pH is measured after various volumes of titrant

have been added to produce a titration curve. The equivalence point can then be read off the

curve.

EXAMPLE:

If 80.0 mL of 0.200 M NaOH are required to reach the equivalence point in a titration of 50.0

mL of hydrofluoric acid, what is the concentration of the hydrofluoric acid?

The neutralization reaction goes to completion:

HF + OH- F

-+ H2O

The number of moles of NaOH added was:

nNaOH= [NaOH] x VNaOH

nNaOH= 0.200 M x 80.0 mL

nNaOH= 16 mmol

Since each NaOH produces 1 OH-, nOH= nNaOH= 16 mmol

From the 1:1 stoichiometry between HF and OH-, nHF= nOH= 16 mmol

So, the concentration of the original hydrofluoric acid solution was:

[HF] = nHF/ VHF

[HF] = 16 mmol / 50 mL

[HF] = 0.320 M

In the same way, knowing the equivalence point can also be used to calculate other unknown

quantities of interest in acid base reactions, such as concentration of titrant or volume of

solution being titrated, provided that enough other information is known to perform the

calculations.

Acid-base indicators


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Certain chemicals have the special property of appearing one colour when in a solution of one

pH and a different colour when in a solution of a different pH. Such chemicals are known as

acid-base indicators, or simply as indicatorsbecause, when a few drops of indicator are

added to a solution, the colour of the solution serves as an indication of its pH.

Most acid-base indicators are simply weak acids with a protonated (acidic) form that is onecolour and a deprotonated (basic) form that is a different colour. Like all weak acids, they

dissocate according to their Kavalue. For a generic indicator HIn,

HIn(aq) H+(aq) + In

-(aq) so Ka= { [H

+][X

-] / [HX] }

Converting this equation into the form of the Henderson-Hasselbalch equation, we get

pH = pKa+ log([In-]/[HIn])

This shows us that the pH of the solution and the pKaof the indicator together determine the

ratio of In-to HIn. Since the two species are different colours, this ratio in turn determines the

overall colour of the solution.

EXAMPLE:

Consider the indicator phenol red, which has a yellow HIn form, a red In-form, and a Kaof

5.0x10-8

. Now imagine that a few drops of this indicator are added to a solution of pH 2.3.

What colour would the solution be?

To figure this out, we need the above equation:


pH = -log(Ka) + log([In-]/[HIn])

2.3 = 7.3 + log([In-]/[HIn])

-5.0 = log([In-]/[HIn])

1.0x10-5

= [In-]/[HIn]

Thus, the ratio of In-to HIn is 1 to 100 000. Since In

-is red and HIn is yellow, there are 100

000 yellow molecules in solution for every red molecule. For coloured species, this is an

enormous ratio, and the red molecules will be undetectable. Thus, the solution will appear

yellow.

EXAMPLE:

Imagine we have a solution of acetic acid of unknown concentration. How could we measure

the concentration of the solution using only indicators?

STEP 1:Use the indicators to estimate the pH of the solution

To do this, we remove a small amount of the solution and add a drop of indicator. Based on

the resulting colour of the solution, we will know whether the pH is below, within or above

the pH range in which the indicator changes colour. We could just do this with severalrandom indicators and hope that they will provide enough information for us. However, we

will use a more systematic approach.


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Since we do not know the pH of our solution, an indicator that changes colour in the middle

of the pH range is a good one to start with since it will split the pH range into roughly equal

parts, each significantly smaller than the whole range. Let's say we add a drop of phenol red

and the solution turns yellow. Since phenol red is yellow in solutions below pH 6.6, we know

that the pH of our acetic acid solution is below 6.6.

Now we can narrow down the pH range with another indicator. A good choice would be one

that will further narrow down the pH range that we know contains the solution's pH. For

example, let's say we add a drop of methyl red to a new sample of the solution and it turns

red. Now we know that the pH is below 4.8, which gives us more information than we had

before. In contrast, if we had used alizarin yellow, it would have turned yellow, telling us that

the pH is below 10.1. Since the phenol red told us that the pH is below 6.6, the alizarin yellow

test gives us no new information.

Using the results of the phenol red and methyl red tests, our third indicator needs to have a

colour-change range even lower than 4.8. Suppose we use thymol blue, and it turns yellow.

This tells us that the pH of the solution is above 2.8 AND that it is below 8.0. (Since thymolblue changes colour twice, the colour tells us where the pH of the solution lies relative to both

of the pKa's of thymol blue.) So, we now know that the pH is between 2.8 and 4.8.

Suppose we now use methyl orange as a final indicator and it turns yellowish orange. This is

an intermediate colour for methyl orange, meaning that the pH is within its colour-change

range of 3.2 to 4.4. Since it is closer to the yellow end of the range than the red end, we can

assume that the pH is in the upper half of the range, namely between 3.8 and 4.4.

Unfortunately, none of our indicators has a range with one endpoint within this range. Thus,

additional tests will not give us any more conclusive information, so we have locallized the

pH as much as possible.

STEP 2:Convert the pH range into a range of H+concentrations:

Using [H+] = 10

-pH, we find that a pH range of 3.8 to 4.4 corresponds to an H

+concentration

range of 4.0x10-5

M to 1.6x10-4

M.

STEP 3:Use an I.C.E. table and the Kaof acetic acid to convert the H+concentration range to

a range of acetic acid concentrations. 4.0x10-5

M:

HC2H3O2 H+ C2H3O2-

INITIAL x 0 0

CHANGE -4.0x10-5

M +4.0x10-5

M +4.0x10-5

M

EQUILIBRIUM (x-4.0x10-5

)M 4.0x10-5

M 4.0x10-5

M

Ka= { [H+][C2H3O2

-] / [HC2H3O2] }

1.8x10-5

= { (4.0x10-5

)(4.0x10-5

) / (x - 4.0x10-5

) }

x = 1.3x10

-4

M = [HC2H3O2]


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Similarly, an H+concentration of 1.6x10

-4M corresponds to an HC2H3O2concentration of

1.5x10-3

M.

So, the indicators have told us that the concentration of the acetic acid solution is between

1.3x10-4

M and 1.5x10-3

M.

Indicators and Titrations

Learning Goal 29

Explain how indicators are used in titrations and be able to choose an appropriate indicator for

a given titration.

Indicators can also be used to track the pH in a titration, usually as a means of detecting the

equivalence point of the titration.

Consider the titration of an acid with a base in which a few drops of indicator have been

added to the acid being titrated. Initially, the pH of the acid will probably be low enough that

the indicator is almost entirely in its acidic (HIn) form. As base is added the pH will increase,

causing the indicator to change to its basic (In-) form, causing a visible colour change. The

point in the titration at which the colour changes is known as the endpointof the titration.

IMPORTANT POINT

The endpoint of a titration is NOT the same thing as the equivalence point. The equivalence

point is a single point defined by the reaction stoichiometry as the point at which the base (oracid) added exactly neutralizes the acid (or base) being titrated. The endpoint is defined by the

choice of indicator as the point at which the colour changes. Depending on how quickly the

colour changes, the endpoint can occur almost instantaneously or be quite wide.

If an appropriate indicator is chosen such that the endpoint of the titration occurs at theequivalence point, then a colour change in the solution being titrated can be used as a signal

that the equivalence point has been reached. The volume of titrant needed to reach the

equivalence point can then be read, allowing us to calculate concentrations, volumes, Ka's etc.

as described in the previous section.

Intuition may suggest that the endpoint of the titration will occur at the equivalence point if

we choose an indicator whose pKais equal to the pH of the equivalence point. If such an

indicator was chosen, the colour change would be half complete at the equivalence point.

Unfortunately, it is very difficult to tell when the colour change is precisely half complete,

making it difficult to precisely identify the equivalence point. Since it is easiest to tell when

the colour first starts to change, we want the equivalence point to occur then.

EXAMPLE:

What is the best pKafor the indicator if a strong acid is being titrated with a strong base? If a

strong base is being titrated with a strong acid?


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For a strong acid-strong base titration, the equivalence point occurs at pH 7.

When an acid is being titrated with a base, the first observable colour change will occur when

the In-to HIn ratio is approximately 1 to 10.

pH = pKa+ log([In-

]/[HIn])7 = pKa+ log(1/10)

7 = pKa- 1

8 = pKa

When a base is being titrated with an acid, the first observable colour change will occur when

the In-to HIn ratio is approximately 10 to 1.


7 = pKa+ log(10)

7 = pKa+ 1

6 = pKa

So, we see that we actually want different indicators for the two titrations.

In practice, the first colour change does not necessarily have to happen right at the

equivalence point. Since we only care about being able to use the colour change to detect

when the equivalence point has been reached, the crucial requirement for choosing a good

indicator is that it begins to change colour after the volume of titrant required to reach the

quivalence point has been added.

For example, a strong acid-strong base has a very sharp equivalence point, meaning that avery large change in pH occurs due to the addition of a very small amount of titrant, often a

single drop. Therefore, any pH in this range will be reached after the addition of the same

number of drops. This means that any indicator that starts to change colour in this range will

signal equally well that the equivalence point has been reached. It also means that the pH

range over which the indicator changes colour will likely be passed in a single drop of titrant,

resulting in a very sharp endpoint.

A titration involving a weak acid or base, however, has a less steep equivalent point with a

less dramatic pH change. Thus, there is a smaller range of pH's that will be passed at the same

time as the pH as the equivalence point, meaning that the chosen indicator must begin to

change colour closer to the equivalence point than in a strong acid-strong base titration, andthat the endpoint will be wider.

Here are the titrations of both a strong acid and a weak acid with a strong base, using methyl

red and thymol blue as possible indicators.


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In the strong acid titration, both indicators begin to change colour at the equivalence point (50

mL of base) so both work equally well. In the weak acid titration, thymol blue changes colour

at the equivalence point, but methyl red begins to change colour after only 15mL of base are

added, which is far from the equivalence point, illustrating the importance of choosing an

appropriate indicator.

In general, the faster the pH changes in the range where the indicator changes colour, the

sharper the endpoint of the titration will be and the more different indicators will be suitable

for the titration.

One limitation of using indicators to find the equivalence point of a titration is that the

approximate pH of the equivalence point must be known in order to choose an indicator thatwill accurately locate the equivalence point. This means that if the approximate pH of the

equivalence point is not known, it will be difficult to locate the equivalence point. However,


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even if we do not know the pH of the equivalence point, we can get around this problem by

using trial and error to choose the best indicator. To do this, we can repeat the titration several

times with several different indicators, taking note of how sharp or wide the endpoints are-

that is, how quickly the colour changes. Since the endpoint is sharpest when it is closest to the

equivalence point, we can simply look for the indicator that gave the sharpest endpoint and

assume that that one will be most accurate for finding the equivalence point.

Composition and Action of Buffered Solutions

Learning Goal 21

Describe how a buffer of a certain pH is made, and how it works to control pH.

Buffered solutionsor buffersare solutions which resist a change in pH when small amountsof acid or base are added.

Buffers contain an acidic species to neutralize OH-ions and a basic species to neutralize H

+

ions. However, these two species must be able to coexist in a solution, without completely

neutralizing each other. Buffers are therefore made of weak acid-base conjugate pairs, such as

HC2H3O2and C2H3O2-.

If we have a weak acid, HX, and its conjugate base, X-, the following equilibrium occurs:

HX (aq) H

+

(aq) + X

-

(aq)

For this reaction,

Ka= { [H+][X

-] / [HX] }

If an amount of H+ions are added, the above reaction will shift to the left. This will cause the

[H+] to decrease, to close to what it was before, and thus the pH will stay fairly constant.

If OH-ions are added they will remove H

+ions to form water, thus increasing the pH.

However, the equilibrium reaction will shift to the right as H+ions are removed. The [H

+] will

therefore remain fairly constant, as will the pH.

The most effective buffering solutions are those which have similar concentrations of HX and

X-, because then the buffer has the capacity to absorb both acid and base, with the same

effectiveness in either direction.

When choosing an appropriate conjugate acid-base pair to form a buffer at a specific pH, the

most effective buffers have the desired pH within 1.0 of the conjugate acid's pKa.

Ways to make a buffer


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Adding a conjugate base to a weak acid Adding a strong acid to a weak base Adding a strong base to a weak acid

Buffering Capacity and pH

The buffering capacityis the amount of acid or base a buffer can accept without the pH

changing appreciably. The greater the amounts of the conjugate acid-base pair, the more

resistant they are to change in pH.

If we solve the acid-dissociation-constant expression for [H+] we get:

[H+] = Ka{ [HX] / [X

-] },

assuming that the amount of added acid or base is less than 5% of the conjugate acid / base

molarity.

We can use this to determine the pH of a buffer:

We first take the negative log of both sides:

-log [H+] = -log [Ka{ [HX] / [X

-] }] = -log Ka- log { [HX] / [X

-] }

We know -log [H+] = pH and -log Ka= pKa.

pH = pKa- log { [HX] / [X-] } = pKa+ log { [X

-] / [HX] }

In general:

pH = pKa+ log { [base] / [acid] }

This is known as the Henderson-Hasselbalch equation.

Additions of Acids or Bases to Buffers

Learning Goal 22

Calculate the change in pH of a simple buffer solution, whose composition is known, caused

by adding small amounts of strong acid or strong base.

EXAMPLE

A 1 L solution containing 0.100 moles of HC2H3O2and 0.100 moles of C2H3O2forms a buffer

solution, pH = 4.74. What is the pH after 0.020 moles of NaOH are added?

STEP 1:Stoichiometry calculation:


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The OH-will react completely with the weak acid HC2H3O2:

HC2H3O2(aq) + OH-(aq) H2O (l) + C2H3O2

-(aq)

HC2H3O2 OH- C2H3O2

-

INITIAL 0.100 mol 0.020 mol 0.100 molCHANGE -0.020 mol -0.020 mol +0.020 mol

FINAL 0.080 mol 0 mol 0.120 mol

STEP 2:Equilibrium calculation:

The solution contains the weak acid and its conjugate base. We shall therefore use their

equilibrium equation, and create an I.C.E. table :

HC2H3O2(aq) H+(aq) + C2H3O2

-(aq)

HC2H3O2 H+ C2H3O2

-

Initial 0.080 M 0 0.120 M

Change -x M +x M +x M

Equilibrium (0.080 -x) M x M (0.120 + x) M

Ka= { [C2H3O2-][H

+] / [HC2H3O2] } = { (0.120 + x)(x) / (0.080 - x) } = 1.8 x 10

-5

1.8 x 10-5

= { (0.120) x / (0.080) }

x = [H+] = 1.2 x 10

-5M

pH = -log (1.2 x 10-5

) = 4.92

Applications of Buffer Calculations

In many systems, such as biological applications, there is extreme sensitivity to minute pH

changes. To study these systems, the pH must be controlled by a buffer that is effective within

given limits. How does one determine what the buffer concentration is?

Learning Goal 23

Calculate the specific amounts of species necessary to make the buffer effective within given

requirements.

EXAMPLE

What must the minimum concentration of HC2H3O3be in a one litre buffer solution of

HC2H3O3-C2H3O3-(pH = 4.74) if the pH changes by less than 0.1 if 0.050 moles of HCl are

added?

STEP 1: Stoichiometry calculation:

The amount of HCl added will react completely with the conjugate base, C2H3O3-:

C2H3O3-+ H

+ HC2H3O3


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The pH of the buffered solution, 4.74, matches the pKaof acetic acid, so from the Henderson-

Hasselbach equation, we know that there are originally equal concentrations of acid and

conjugate base, which we denote as x:

For a complete derivation

HC2H3O3 H+ C2H3O3

-

INITIAL x mol 0.050 mol x mol

CHANGE +0.050 mol -0.050 mol -0.050 mol

FINAL (x+0.050) mol 0 mol (x-0.050) mol

Because there is one litre of solution, [H2H3O3] = (x + 0.050)M and [C2H3O3-] = (x - 0.050)M.

STEP 2: Calculate the equilibrium concentration that will produce the desired pH:

Having added acid, the pH will decrease. However, the tolerance given is 0.1, so the lowest

pH that the final solution can have is 4.64.

The hydronium concentration is thus:

[H+] = 10

-pH

[H+] = 10

-4.64= 2.29 x 10

-5M

STEP 3: Equilibrium calculation:

I.C.E. table:

HC2H3O3 H++ C2H3O3

-

HC2H3O3 H+ C2H3O3

-

INITIAL (x+0.050)M 0M (x-0.050)M

CHANGE -2.29 x 10-5M

+2.29 x 10-5M

+2.29 x 10-5M

EQUILIBRIUM (x+0.050)M 2.29 x 10-5M

(x-0.050)M

Ka= { [H+][C2H3O2

-] } / [HC2H3O2] = { (2.29x10

-5)(x-0.050) } / (x+0.050) = 1.8x10

-5

x = [HC2H3O2] = 0.417M

Thus, the molarity of both [HC2H3O2] and [C2H3O2-] must be equal to or greater than 0.417M

to produce the required effectiveness.


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4.1.3. Buffer solutions EUROPEAN PHARMACOPOEIA 5.0

01/2005:40103

4.1.3. BUFFER SOLUTIONS

Buffered acetone solution. 4000100.

Dissolve 8.15 g ofsodium acetate R and 42 g ofsodiumchloride Rin water R, add 68 ml of0.1 M hydrochloric acid

and 150 ml ofacetone R and dilute to 500 ml with water R.

Buffer solution pH 2.0. 4000200.

Dissolve 6.57 g ofpotassium chloride R in water Rand add119.0 ml of0.1 M hydrochloric acid. Dilute to 1000.0 mlwith water R.

Phosphate buffer solution pH 2.0. 4007900.

Dissolve 8.95 g ofdisodium hydrogen phosphate Rand3.40 g ofpotassium dihydrogen phosphate Rin water Rand dilute to 1000.0 ml with the same solvent. If necessaryadjust the pH (2.2.3) withphosphoric acid R.

Sulphate buffer solution pH 2.0. 4008900.

Dissolve 132.1 g ofammonium sulphate Rin water Rand dilute to 500.0 ml with the same solvent (Solution I).Carefully and with constant cooling stir 14 ml of sulphuricacid R into about 400 ml ofwater R ; allow to cool and diluteto 500.0 ml withwater R(Solution II). Mix equal volumes ofsolutions I and II. Adjust the pH (2.2.3) if necessary.


Mix of 6.7 ml ofphosphoric acid R with 50.0 ml of a 4 percent solution ofdilute sodium hydroxide solution Randdilute to 1000.0 ml withwater R.


Dissolve 100 g ofpotassium dihydrogen phosphate R in

800 ml ofwater R ; adjust to pH 2.5 (2.2.3) withhydrochloricacid Rand dilute to 1000.0 ml with water R.

Buffer solution pH 2.5 R1. 4000400.

To 4.9 g ofdilute phosphoric acid Radd 250 ml ofwater R.Adjust the pH (2.2.3) withdilute sodium hydroxide

solution R and dilute to 500.0 ml with water R.


Dissolve 7.8 g ofsodium dihydrogen phosphate R in 900 mlofwater R, adjust to pH 2.8 (2.2.3) withphosphoric acid Rand dilute to 1000 ml with the same solvent.


Dissolve 21.0 g ofcitric acid R in 200 ml of1 M sodiumhydroxideand dilute to 1000 ml withwater R. Dilute 40.3 mlof this solution to 100.0 ml with0.1 M hydrochloric acid.


Mix 0.7 ml ofphosphoric acid Rwith 100 ml ofwater R.Dilute to 900 ml with the same solvent. Adjust to pH 3.0(2.2.3) withstrong sodium hydroxide solution R and diluteto 1000 ml withwater R.

0.1 M Phosphate buffer solution pH 3.0. 4011500.

Dissolve 12.0 g ofanhydrous sodium dihydrogenphosphate Rin water R, adjust the pH (2.2.3) withdilutephosphoric acid R1 and dilute to 1000 ml with water R.

Phosphate buffer solution pH 3.0 R1. 4010000.

Dissolve 3.40 g ofpotassium dihydrogen phosphate R in900 ml ofwater R. Adjust to pH 3.0 (2.2.3) withphosphoricacid Rand dilute to 1000.0 ml with water R.


To 900 ml of a 4 g/l solution ofsodium dihydrogenphosphate R, add 100 ml of a 2.5 g/l solution ofphosphoricacid R. Adjust the pH (2.2.3) if necessary.


Adjust a 35.8 g/l solution ofdisodium hydrogenphosphate Rto pH 3.2 (2.2.3) withdilute phosphoric acid R.Dilute 100.0 ml of the solution to 2000.0 ml withwater R.


Dissolve 25.0 g ofammonium acetate Rin 25 ml ofwater Rand add 38.0 ml ofhydrochloric acid R1. Adjust the pH(2.2.3) if necessary withdilute hydrochloric acid R or diluteammonia R1. Dilute to 100.0 ml with water R.


Dissolve 68.0 g ofpotassium dihydrogen phosphate Rinwater Rand dilute to 1000.0 ml with the same solvent.Adjust the pH (2.2.3) withphosphoric acid R.


To 250.0 ml of0.2 M potassium hydrogen phthalate R add11.94 ml of0.2 M hydrochloric acid. Dilute to 1000.0 mlwith water R.


To 15.0 ml ofacetic acid R add 60 ml ofalcohol Rand20 ml ofwater R. Adjust to pH 3.7 (2.2.3) by the addition ofammonia R. Dilute to 100.0 ml with water R.

Buffered copper sulphate solution pH 4.0. 4001000.

Dissolve 0.25 g ofcopper sulphate Rand 4.5 g ofammoniumacetate Rin dilute acetic acid R and dilute to 100.0 ml withthe same solvent.

Acetate buffer solution pH 4.4. 4001100.Dissolve 136 g ofsodium acetate Rand 77 g ofammoniumacetate Rin water R and dilute to 1000.0 ml with the samesolvent; add 250.0 ml ofglacial acetic acid R and mix.

Phthalate buffer solution pH 4.4. 4001200.

Dissolve 2.042 g ofpotassium hydrogen phthalate R in50 ml ofwater R, add 7.5 ml of0.2 M sodium hydroxideanddilute to 200.0 ml withwater R.

Acetate buffer solution pH 4.5. 4012500.

Dissolve 77.1 g ofammonium acetate Rin water R. Add70 ml ofglacial acetic acid Rand dilute to 1000.0 ml withwater R.


Dissolve 6.80 g ofpotassium dihydrogen phosphate Rin1000.0 ml ofwater R. The pH (2.2.3) of the solution is 4.5.

Sodium acetate buffer solution pH 4.5. 4010100.

Dissolve 63 g ofanhydrous sodium acetate Rin water R,add 90 ml acetic acid R and adjust to pH 4.5, and dilute to1000 ml withwater R.


Dissolve 5.4 g ofsodium acetate Rin 50 ml ofwater R, add2.4 g ofglacial acetic acid R and dilute to 100.0 ml withwater R. Adjust the pH (2.2.3) if necessary.

Succinate buffer solution pH 4.6. 4001500.

Disssolve 11.8 g ofsuccinic acid R in a mixture of 600 mlofwater Rand 82 ml of1 M sodium hydroxideand diluteto 1000.0 ml withwater R.

430 See the information section on general monographs (cover pages)


17/63

EUROPEAN PHARMACOPOEIA 5.0 4.1.3. Buffer solutions


Dissolve 136.1 g ofsodium acetate Rin 500 ml ofwater R.Mix 250 ml of this solution with 250 ml ofdilute aceticacid R. Shake twice with a freshly prepared, filtered, 0.1 g/lsolution ofdithizone Rin chloroform R. Shake withcarbontetrachloride Runtil the extract is colourless. Filter theaqueous layer to remove traces of carbon tetrachloride.

Acetate buffer solution pH 5.0. 4009100.To 120 ml of a 6 g/l solution ofglacial acetic acid R add100 ml of0.1 M potassium hydroxide and about 250 ml ofwater R. Mix. Adjust the pH to 5.0 with a 6 g/l solution ofacetic acid Ror with0.1 M potassium hydroxide and diluteto 1000.0 ml with water R.

Citrate buffer solution pH 5.0. 4010700.

Prepare a solution containing 20.1 g/l ofcitric acid Rand8.0 g/l ofsodium hydroxide R. Adjust the pH with dilutehydrochloric acid R.


Dissolve 2.72 g ofpotassium dihydrogen phosphate R in800 ml ofwater R. Adjust the pH (2.2.3) with1 M potassiumhydroxideand dilute to 1000 ml with water R.


Dissolve 1.02 g ofpotassium hydrogen phthalate R in30.0 ml of0.1 M sodium hydroxide. Dilute to 100.0 ml withwater R.


Mix appropriate volumes of a 23.99 g/l solution ofdisodiumhydrogen phosphate Rwith a 9.12 g/l solution ofsodiumdihydrogen phosphate monohydrate R to obtain pH 5.4(2.2.3).

Acetate-edetate buffer solution pH 5.5. 4001900.Dissolve 250 g ofammonium acetate R and 15 g sodiumedetate Rin 400 ml ofwater Rand add 125 ml ofglacialacetic acid R.


Dissolve 54.4 g ofsodium acetate R in 50 ml ofwater R,heating to 35 C if necessary. After cooling, slowly add 10 mlofanhydrous acetic acid R. Shake and dilute to 100.0 mlwithwater R.


Solution I. Dissolve 13.61 g ofpotassium dihydrogenphosphate R in water Rand dilute to 1000.0 ml with thesame solvent.

Solution II. Dissolve 35.81 g ofdisodium hydrogenphosphate R in water Rand dilute to 1000.0 ml with thesame solvent.

Mix 96.4 ml of solution I and 3.6 ml of solution II.

Phosphate-citrate buffer solution pH 5.5. 4008700.

Mix 56.85 ml of a 28.4 g/l solution ofanhydrous disodiumhydrogen phosphate R and 43.15 ml of a 21 g/l solutionofcitric acid R.


Solution I. Dissolve 0.908 g ofpotassium dihydrogen

phosphate R in water Rand dilute to 100.0 ml with thesame solvent.

Solution II. Dissolve 1.161 g ofdipotassium hydrogenphosphate R in water Rand dilute to 100.0 ml with thesame solvent.

Mix 94.4 ml of solution I and 5.6 ml of solution II. Ifnecessary, adjust to pH 5.6 (2.2.3) using solution I orsolution II.


Dissolve 1.19g ofdisodium hydrogen phosphate dihydrate Rand 8.25 g ofpotassium dihydrogen phosphate R inwater Rand dilute to 1000.0 ml with the same solvent.


Dissolve 100 g ofammonium acetate Rin 300 ml ofwater R,add 4.1 ml ofglacial acetic acid R, adjust the pH (2.2.3) ifnecessary usingammonia Ror acetic acid Rand dilute to500.0 ml withwater R.

Diethylammonium phosphate buffer solution pH 6.0.4002300.

Dilute 68 ml ofphosphoric acid Rto 500 ml with water R.To 25 ml of this solution add 450 ml ofwater Rand 6 ml ofdiethylamine R, adjust to pH 6 0.05 (2.2.3), if necessary,usingdiethylamine Ror phosphoric acid R and dilute to500.0 ml withwater R.


Mix 63.2 ml of a 71.5 g/l solution ofdisodium hydrogenphosphate Rand 36.8 ml of a 21 g/l solution ofcitric acid R.


Dissolve 6.8 g ofsodium dihydrogen phosphate R inwater Rand dilute to 1000.0 ml withwater R. Adjust the pH (2.2.3)withstrong sodium hydroxide solution R.


To 250.0 ml of0.2 M potassium dihydrogen phosphate Radd 28.5 ml of0.2 M sodium hydroxideand dilute to

1000.0 ml withwater R.


Dissolve 2.5 g ofdisodium hydrogen phosphate R, 2.5 gofsodium dihydrogen phosphate Rand 8.2 g ofsodiumchloride Rin 950 ml ofwater R. Adjust the pH (2.2.3) ofthe solution to 6.4 with 1 M sodium hydroxide or 1 Mhydrochloric acid, if necessary. Dilute to 1000.0 ml withwater R.

0.5 M Phthalate buffer solution pH 6.4. 4009200.

Dissolve 100 g ofpotassium hydrogen phthalate Rinwater Rand dilute to 1000.0 ml with the same solvent.Adjust the pH (2.2.3) if necessary, usingstrong sodium

hydroxide solution R.


Dissolve 60.5 g ofdisodium hydrogen phosphate R and46 g ofpotassium dihydrogen phosphate R in water R. Add100 ml of0.02 M sodium edetateand 20 mg ofmercuricchloride Rand dilute to 1000.0 ml with water R.

Imidazole buffer solution pH 6.5. 4003000.

Dissolve 6.81 g ofimidazole R, 1.23 g ofmagnesiumsulphate R and 0.73 g ofcalcium sulphate R in 752 ml of0.1 M hydrochloric acid. Adjust the pH (2.2.3) if necessaryand dilute to 1000.0 ml with water R.

0.1 M phosphate buffer solution pH 6.5. 4010800.Dissolve 13.80 g ofsodium dihydrogen phosphatemonohydrate Rin 900 ml ofdistilled water R. Adjust thepH (2.2.3) using a 400 g/l solution ofsodium hydroxide R.Dilute to 1000 ml withdistilled water R.

General Notices (1) apply to all monographs and other texts 431


18/63



To 250.0 ml of0.2 M potassium dihydrogen phosphate Radd 89.0 ml of0.2 M sodium hydroxide. Dilute to 1000.0 mlwithwater R.

Phosphate buffered saline pH 6.8. 4003200.

Dissolve 1.0 g ofpotassium dihydrogen phosphate R, 2.0 gofdipotassium hydrogen phosphate R and 8.5 g ofsodium

chloride Rin 900 ml ofwater R, adjust the pH (2.2.3) ifnecessary and dilute to 1000.0 ml with the same solvent.


Mix 77.3 ml of a 71.5 g/l solution ofdisodium hydrogenphosphate Rwith 22.7 ml of a 21 g/l solution ofcitric acid R.


To 51.0 ml of a 27.2 g/l solution ofpotassium dihydrogenphosphate R add 49.0 ml of a 71.6 g/l solution ofdisodiumhydrogen phosphate R. Adjust the pH (2.2.3) if necessary.

Storage : at 2 C to 8 C.

1 M tris-hydrochloride buffer solution pH 6.8. 4009300.

Dissolve 60.6 g oftris(hydroxymethyl)aminomethane R in400 ml ofwater R. Adjust the pH (2.2.3) withhydrochloricacid Rand dilute to 500.0 ml withwater R.


To 1000 ml of a solution containing 18 g/l ofdisodiumhydrogen phosphate Rand 23 g/l ofsodium chloride R addsufficient (about 280 ml) of a solution containing 7.8 g/lofsodium dihydrogen phosphate Rand 23 g/l ofsodiumchloride Rto adjust the pH (2.2.3). Dissolve in the solutionsufficientsodium azide R to give a 0.2 g/l solution.

Maleate buffer solution pH 7.0. 4003600.

Dissolve 10.0 g ofsodium chloride R, 6.06 g of

tris(hydroxymethyl)aminomethane Rand 4.90 g ofmaleicanhydride Rin 900 ml ofwater R. Adjust the pH (2.2.3)using a 170 g/l solution ofsodium hydroxide R. Dilute to1000.0 ml withwater R.

Storage : at 2 C to 8 C; use within 3 days.


Mix 1 volume of0.063 M phosphate buffer solution pH 7.0 Rwith 1.5 volumes ofwater R.


Dissolve 5.2 g ofdipotassium hydrogen phosphate R in900 ml ofwater for chromatography R. Adjust the solutionto pH 7.0 0.1 usingphosphoric acid Rand dilute to1000 ml withwater for chromatography R.


Mix 34 ml ofwater Rand 100 ml of0.067 M phosphatebuffer solution pH 7.0 R.


Dissolve 5.18 g ofanhydrous disodium hydrogenphosphate R and 3.65 g ofsodium dihydrogen phosphatemonohydrate Rin 950 ml ofwater Rand adjust the pH(2.2.3) withphosphoric acid R ; dilute to 1000.0 ml withwater R.


Solution I. Dissolve 0.908 g ofpotassium dihydrogen

phosphate R in water Rand dilute to 100.0 ml with thesame solvent.

Solution II. Dissolve 2.38 g ofdisodium hydrogenphosphate R in water Rand dilute to 100.0 ml with thesame solvent.

Mix 38.9 ml of solution I and 61.1 ml of solution II. Adjustthe pH (2.2.3) if necessary.


Dissolve 1.361 g ofpotassium dihydrogen phosphate Rinwater Rand dilute to 100.0 ml with the same solvent. Adjustthe pH (2.2.3) using a 35 g/l solution ofdisodium hydrogen

phosphate R.




Mix 250.0 ml of0.2 M potassium dihydrogen phosphate Rand 148.2 ml of a 8 g/l solution ofsodium hydroxide R,adjust the pH (2.2.3) if necessary. Dilute to 1000.0 ml withwater R.


Mix 50.0 ml of a 136 g/l solution ofpotassium dihydrogen

phosphate Rwith 29.5 ml of1 M sodium hydroxide anddilute to 100.0 ml with water R. Adjust the pH (2.2.3) to7.0 0.1.


Dissolve 5 g ofpotassium dihydrogen phosphate R and 11 gofdipotassium hydrogen phosphate Rin 900 ml ofwater R.Adjust to pH 7.0 (2.2.3) withdilute phosphoric acid R ordilute sodium hydroxide solution R. Dilute to 1000 ml withwater Rand mix.


Dissolve 28.4 g ofanhydrous disodium hydrogenphosphate Rand 18.2 g ofpotassium dihydrogen

phosphate R in water Rand dilute to 500 ml with the samesolvent.


Dissolve 28.4 g ofanhydrous disodium hydrogenphosphate Rin 800 ml ofwater R. Adjust the pH (2.2.3)using a 30 per centm/m solution ofphosphoric acid Randdilute to 1000 ml with water R.

Tetrabutylammonium buffer solution pH 7.0. 4010900.

Dissolve 6.16 g ofammonium acetate Rin a mixture of 15 mloftetrabutylammonium hydroxide solution (400 g/l) R and185 ml ofwater R. Adjust the pH (2.2.3) withnitric acid R.

Buffered salt solution pH 7.2. 4004300.Dissolve inwater R 8.0 g ofsodium chloride R, 0.2 gofpotassium chloride R, 0.1 g ofanhydrous calciumchloride R, 0.1 g ofmagnesium chloride R, 3.18 g ofdisodium hydrogen phosphate R and 0.2 g ofpotassiumdihydrogen phosphate R and dilute to 1000.0 ml withwater R.


To 250.0 ml of0.2 M potassium dihydrogen phosphate Radd 175.0 ml of0.2 M sodium hydroxide. Dilute to 1000.0 mlwithwater R. Adjust the pH (2.2.3) if necessary.

Phosphate-albumin buffered saline pH 7.2. 4004400.

Dissolve 10.75 g ofdisodium hydrogen phosphate R, 7.6 gofsodium chloride R and 10 g ofbovine albumin R inwater Rand dilute to 1000.0 ml with the same solvent.Immediately before use adjust the pH (2.2.3) usingdilute

sodium hydroxide solution R or dilute phosphoric acid R.



19/63

EUROPEAN PHARMACOPOEIA 5.0 4.1.3. Buffer solutions

Phosphate-albumin buffered saline pH 7.2 R1. 4009600.

Dissolve 10.75 g ofdisodium hydrogen phosphate R, 7.6 gofsodium chloride Rand 1 g ofbovine albumin R inwater Rand dilute to 1000.0 ml with the same solvent.Immediately before use adjust the pH (2.2.3) usingdilute

sodium hydroxide solution R or dilute phosphoric acid R.



Imidazole buffer solution pH 7.3. 4004500.

Dissolve 3.4 g ofimidazole Rand 5.8 g ofsodium chloride Rinwater R, add 18.6 ml of1 M hydrochloric acidand diluteto 1000.0 ml withwater R. Adjust the pH (2.2.3) if necessary.

Barbital buffer solution pH 7.4. 4004700.

Mix 50 ml of a solution in water Rcontaining 19.44 g/l ofsodium acetate R and 29.46 g/l ofbarbital sodium Rwith50.5 ml of0.1 M hydrochloric acid, add 20 ml of an 85 g/lofsodium chloride R and dilute to 250 ml withwater R.


Dissolve 0.6 g ofpotassium dihydrogen phosphate R, 6.4 gofdisodium hydrogen phosphate Rand 5.85 g ofsodiumchloride Rin water R, and dilute to 1000.0 ml with the samesolvent. Adjust the pH (2.2.3) if necessary.

Phosphate buffered saline pH 7.4. 4005000.

Dissolve 2.38 g ofdisodium hydrogen phosphate R, 0.19 gofpotassium dihydrogen phosphate R and 8.0 g ofsodiumchloride Rin water. Dilute to 1000.0 ml with the samesolvent. Adjust the pH (2.2.3) if necessary.


Add 250.0 ml of0.2 M potassium dihydrogen phosphate Rto 393.4 ml of0.1 M sodium hydroxide.

Tris(hydroxymethyl)aminomethane buffer solution pH 7.4.4012100.

Dissolve 30.3 g oftris(hydroxymethyl)aminomethane Rin approximately 200 ml ofwater R. Add 183 ml of1 Mhydrochloric acid. Dilute to 500.0 ml with water R.Note:the pH is 7.7-7.8 at room temperature and 7.4 at 37 C. This

solution is stable for several months at 4 C.

Tris(hydroxymethyl)aminomethane sodium chloride buffersolution pH 7.4. 4004900.

Dissolve 6.08 g oftris(hydroxymethyl)aminomethane R,8.77 g ofsodium chloride R in 500 ml ofdistilled water R.Add 10.0 g ofbovine albumin R. Adjust the pH (2.2.3) usinghydrochloric acid R. Dilute to 1000.0 ml with distilledwater R.

Tris(hydroxymethyl)aminomethane sodium chloride buffersolution pH 7.4 R1. 4012200.

Dissolve 0.1 g ofbovine albumin Rin a mixture containing2 ml oftris(hydroxymethyl)aminomethane buffer solution

pH 7.4 R and 50 ml of a 5.84 mg/ml solution ofsodiumchloride R. Dilute to 100.0 ml with water R.

Borate buffer solution pH 7.5. 4005200.

Dissolve 2.5 g ofsodium chloride R, 2.85 g ofdisodiumtetraborate Rand 10.5 g ofboric acid R in water Randdilute to 1000.0 ml with the same solvent. Adjust the pH(2.2.3) if necessary.

Storage : at 2 C to 8 C.

Buffer (HEPES) solution pH 7.5. 4009700.

Dissolve 2.38 g of2-[4-(2-hydroxyethyl)piperazin-1-yl]ethanesulphonic acid Rin about 90 ml ofwater R. Adjustthe pH to 7.5 with sodium hydroxide solution R. Dilute to100 ml withwater R.


Dissolve 27.22 g ofpotassium dihydrogen phosphate R in

930 ml ofwater R, adjust to pH 7.5 (2.2.3) with a 300 g/lsolution ofpotassium hydroxide Rand dilute to 1000.0 mlwith water R.


Solution I. Dissolve 119.31 g ofdisodium hydrogenphosphate Rin water Rand dilute to 1000.0 ml with thesame solvent.

Solution II. Dissolve 45.36 g ofpotassium dihydrogenphosphate Rin water Rand dilute to 1000.0 ml with thesame solvent.

Mix 85 ml of solution I and 15 ml of solution II. Adjust thepH (2.2.3) if necessary.

0.05 M Tris-hydrochloride buffer solution pH 7.5.4005600.

Dissolve 6.057 g oftris(hydroxymethyl)aminomethane Rinwater Rand adjust the pH (2.2.3) withhydrochloric acid R.Dilute to 1000.0 ml withwater R.

Tris(hydroxymethyl)aminomethane buffer solution pH 7.5.4005500.

Dissolve 7.27 g oftris(hydroxymethyl)aminomethane Rand5.27 g ofsodium chloride Rin water R, and adjust the pH(2.2.3) if necessary. Dilute to 1000.0 ml with water R.

Sodium citrate buffer solution pH 7.8 (0.034 M sodiumcitrate, 0.101 M sodium chloride). 4009800.

Dissolve 10.0 g ofsodium citrate R and 5.90 g ofsodiumchloride Rin 900 ml ofwater R. Adjust the pH (2.2.3) byaddition ofhydrochloric acid Rand dilute to 1000 ml withwater R.

0.0015 M Borate buffer solution pH 8.0. 4006000.

Dissolve 0.572 g ofdisodium tetraborate R and 2.94 g ofcalcium chloride Rin 800 ml ofwater R. Adjust the pH(2.2.3) with1 M hydrochloric acid. Dilute to 1000.0 ml withwater R.


To 50.0 ml of0.2 M potassium dihydrogen phosphate Radd 46.8 ml of0.2 M sodium hydroxide. Dilute to 200.0 mlwith water R.


Dissolve 20 g ofdipotassium hydrogen phosphate R in900 ml ofwater R. Adjust the pH (2.2.3) withphosphoricacid R. Dilute to 1000 ml with water R.


To 50.0 ml of0.2 M potassium dihydrogen phosphate Radd 46.8 ml of0.2 M sodium hydroxide. Dilute to 500.0 mlwith water R.


Dissolve 0.523 g ofpotassium dihydrogen phosphate R and16.73 g ofdipotassium hydrogen phosphate R in water Rand dilute to 1000.0 ml with the same solvent.

1 M Phosphate buffer solution pH 8.0. 4007800.

Dissolve 136.1 g ofpotassium dihydrogen phosphate R inwater R, adjust the pH (2.2.3) with1 M sodium hydroxide.Dilute to 1000.0 ml withwater R.



20/63


Tris-hydrochloride buffer solution pH 8.0. 4012300.

Dissolve 1.21 g oftris(hydroxymethyl)aminomethane Rand29.4 mg ofcalcium chloride Rin water R. Adjust the pH(2.2.3) with1 M hydrochloric acidand dilute to 100.0 mlwithwater R.

Tris(hydroxymethyl)aminomethane buffer solution pH 8.1.

4006200.Dissolve 0.294 g ofcalcium chloride Rin 40 ml oftris(hydroxymethyl)aminomethane solution Rand adjustthe pH (2.2.3) with1 M hydrochloric acid. Dilute to 100.0 mlwith water R.

Tris-glycine buffer solution pH 8.3. 4006300.

Dissolve 6.0 g oftris(hydroxymethyl)aminomethane Rand 28.8 g ofglycine Rin water Rand dilute to 1000.0 mlwith the same solvent. Dilute 1 volume to 10 volumes withwater R immediately before use.

Tris-hydrochloride buffer solution pH 8.3. 4011800.

Dissolve 9.0 g oftris(hydroxymethyl)aminomethane Rin 2.9 litres ofwater R. Adjust the pH (2.2.3) with1 Mhydrochloric acid. Adjust the volume to 3 litres withwater R.

Barbital buffer solution pH 8.4. 4006400.

Dissolve 8.25 g ofbarbital sodium Rin water Rand diluteto 1000.0 ml with the same solvent.

Tris-EDTA BSA buffer solution pH 8.4. 4006500.

Dissolve 6.1 g oftris(hydroxymethyl)aminomethane R, 2.8 gofsodium edetate R, 10.2 g ofsodium chloride Rand 10 g ofbovine albumin Rin water R, adjust to pH 8.4 (2.2.3) using1 M hydrochloric acidand dilute to 1000.0 ml with water R.

Tris(hydroxymethyl)aminomethane EDTA buffer solutionpH 8.4. 4006600.

Dissolve 5.12 g ofsodium chloride R, 3.03 g oftris(hydroxymethyl)aminomethane Rand 1.40 g ofsodiumedetate Rin 250 ml ofdistilled water R. Adjust the pH(2.2.3) to 8.4 usinghydrochloric acid R. Dilute to 500.0 mlwith distilled water R.

Tris acetate buffer solution pH 8.5. 4006700.

Dissolve 0.294 g ofcalcium chloride R and 12.11 g oftris(hydroxymethyl)aminomethane Rin water R. Adjustthe pH (2.2.3) withacetic acid R. Dilute to 1000.0 ml withwater R.

Barbital buffer solution pH 8.6 R1. 4006900.

Dissolve inwater R 1.38 g ofbarbital R, 8.76 g ofbarbitalsodium R and 0.38 g ofcalcium lactate Rand dilute to1000.0 ml with the same solvent.

1.5 M tris-hydrochloride buffer solution pH 8.8. 4009900.

Dissolve 90.8 g oftris(hydroxymethyl)aminomethane R in400 ml ofwater R. Adjust the pH (2.2.3) withhydrochloricacid Rand dilute to 500.0 ml withwater R.

Buffer (phosphate) solution pH 9.0. 4008300.

Dissolve 1.74 g ofpotassium dihydrogen phosphate Rin80 ml ofwater R, adjust the pH (2.2.3) with1 M potassiumhydroxideand dilute to 100.0 ml with water R.


Solution I. Dissolve 6.18 g ofboric acid Rin0.1 M potassiumchloride Rand dilute to 1000.0 ml with the same solvent.

Solution II.0.1 M sodium hydroxide.

Mix 1000.0 ml of solution I and 420.0 ml of solution II.


Dissolve 6.20 g ofboric acid Rin 500 ml ofwater Randadjust the pH (2.2.3) with1 M sodium hydroxide (about41.5 ml). Dilute to 1000.0 ml with water R.

Ammonium chloride buffer solution pH 9.5. 4007200.

Dissolve 33.5 g ofammonium chloride Rin 150 ml ofwater R, add 42.0 ml ofconcentrated ammonia Rand diluteto 250.0 ml with water R.

Storage : in a polyethylene container.


Dissolve 5.4 g ofammonium chloride Rin 20 ml ofwater R,add 35.0 ml ofammonia R and dilute to 100.0 ml withwater R.

Diethanolamine buffer solution pH 10.0. 4007500.

Dissolve 96.4 g ofdiethanolamine Rin water Rand diluteto 400 ml with the same solvent. Add 0.5 ml of an 186 g/lsolution ofmagnesium chloride Rand adjust the pH (2.2.3)with1 M hydrochloric acid. Dilute to 500.0 ml withwater R.

0.1 M Ammonium carbonate buffer solution pH 10.3.4011900.

Dissolve 7.91 g ofammonium carbonate R in 800 ml ofwater R. Adjust the pH (2.2.3) withdilute sodium hydroxide

solution R. Dilute to 1000.0 ml with water R.


Dissolve 70 g ofammonium chloride Rin 200 ml ofwater R,add 330 ml ofconcentrated ammonia Rand dilute to1000.0 ml withwater R. If necessary, adjust to pH 10.4 withammonia R.

Borate buffer solution pH 10.4. 4011100.

Dissolve 24.64 g ofboric acid R in 900 ml ofdistilledwater R. Adjust the pH (2.2.3) using a 400 g/l solutionofsodium hydroxide R. Dilute to 1000 ml with distilledwater R.


Dissolve 6.75 g ofammonium chloride R in ammonia Rand dilute to 100.0 ml with the same solvent.

Total-ionic-strength-adjustment buffer. 4007700.

Dissolve 58.5 g ofsodium chloride R, 57.0 ml ofglacialacetic acid R, 61.5 g ofsodium acetate Rand 5.0 g ofcyclohexylene-dinitrilotetra-acetic acid Rin water Randdilute to 500.0 ml with the same solvent. Adjust to pH 5.0to 5.5 with a 335 g/l solution ofsodium hydroxide R anddilute to 1000.0 ml with distilled water R.

Total-ionic-strength-adjustment buffer R1. 4008800.

Solution (a). Dissolve 210 g ofcitric acid R in 400 ml of

distilled water R. Adjust to pH 7.0 (2.2.3) withconcentratedammonia R. Dilute to 1000.0 ml withdistilled water R.

Solution (b). Dissolve 132 g ofammonium phosphate Rindistilled water Rand dilute to 1000.0 ml with the samesolvent.



21/63

EUROPEAN PHARMACOPOEIA 5.0 4.2.2. Volumetric solutions

Solution (c). To a suspension of 292 g of(ethylenedinitrilo)tetra-acetic acid R in about 500 mlofdistilled water R, add about 200 ml ofconcentratedammonia Rto dissolve. Adjust the pH to 6 to 7 (2.2.3) withconcentrated ammonia R. Dilute to 1000.0 ml with distilledwater R.

Mix equal volumes of solution (a), (b), and (c) and adjust topH 7.5 withconcentrated ammonia R.

4.2. VOLUMETRIC ANALYSIS

01/2005:40201

4.2.1. PRIMARY STANDARDS FORVOLUMETRIC SOLUTIONS

Primary standards for volumetric solutions are indicatedby the suffix RV. Primary standards of suitable quality maybe obtained from commercial sources or prepared by thefollowing methods.

Benzoic acid. C7H6O2. (Mr122.1). 2000200. [65-85-0].

Sublimebenzoic acid Rin a suitable apparatus.

Potassium bromate. KBrO3. (Mr167.0). 2000300.[7758-01-2].

Crystallise potassium bromate R from boiling water R.Collect the crystals and dry to constant mass at 180 C.

Potassium hydrogen phthalate. C8H5KO4. (Mr204.2).2000400. [877-24-7].

Recrystallise potassium hydrogen phthalate R from boilingwater R, collect the crystals at a temperature above 35 Cand dry to constant mass at 110 C.

Sodium carbonate. Na2CO3. (Mr106.0). 2000500.[497-19-8].

Filter at room temperature a saturated solution ofsodiumcarbonate R. Introduce slowly into the filtrate a stream ofcarbon dioxide R with constant cooling and stirring. Afterabout 2 h, collect the precipitate on a sintered-glass filter.Wash the filter with icedwater Rcontaining carbon dioxide.After drying at 100 C to 105 C, heat to constant mass at270 C to 300 C, stirring from time to time.

Sodium chloride. NaCl. (Mr58.44). 2000600. [7647-14-5].

To 1 volume of a saturated solution ofsodium chloride Radd 2 volumes ofhydrochloric acid R. Collect the crystalsformed and wash with hydrochloric acid R1. Remove the

hydrochloric acid by heating on a water-bath and dry thecrystals to constant mass at 300 C.

Sulphanilic acid. C6H7NO3S. (Mr 173.2). 2000700.[121-57-3].

Recrystallise sulphanilic acid Rfrom boilingwater R. Filterand dry to constant mass at 100 C to 105 C.

Zinc. Zn. (Mr65.4). 2000800. [7440-66-6].

Use a quality containing not less than 99.9 per cent of Zn.

01/2005:40202

4.2.2. VOLUMETRIC SOLUTIONSVolumetric solutions are prepared according to the usualchemical analytical methods. The accuracy of the apparatusused is verified to ensure that it is appropriate for theintended use.

The concentration of volumetric solutions is indicatedin terms of molarity. Molarity expresses, as the numberof moles, the amount of substance dissolved in 1 litre ofsolution. A solution which containsxmoles of substance perlitre is said to be xM.

Volumetric solutions do not differ from the prescribedstrength by more than 10 per cent. The molarity of thevolumetric solutions is determined by an appropriate numberof titrations. The repeatability does not exceed 0.2 per cent(relative standard deviation).

Volumetric solutions are standardised by the methodsdescribed below. When a volumetric solution is to beused in an assay in which the end-point is determined byan electrochemical process (for example, amperometryor potentiometry) the solution is standardised by thesame method. The composition of the medium in which avolumetric solution is standardised should be the same asthat in which it is to be used.

Solutions more dilute than those described are obtained bydiluting the latter with carbon dioxide-free water R. Thecorrection factors of these solutions are the same as those

from which the dilutions were prepared.0.1 M Acetic acid. 3008900.

Dilute 6.0 g ofglacial acetic acid R to 1000.0 ml withwater R.

Standardisation. To 25.0 ml of acetic acid add 0.5 ml ofphenolphthalein solution Rand titrate with 0.1 M sodiumhydroxide.

0.1 M Ammonium and cerium nitrate. 3000100.

Shake for 2 min a solution containing 56 ml of sulphuricacid R and 54.82 g ofammonium and cerium nitrate R, addfive successive quantities, each of 100 ml, ofwater R, shakingafter each addition. Dilute the clear solution to 1000.0 mlwithwater R. Standardise the solution after 10 days.

Standardisation. To 25.0 ml of the ammonium and ceriumnitrate solution add 2.0 g ofpotassium iodide Rand150 ml ofwater R. Titrate immediately with0.1 M sodiumthiosulphate, using 1 ml ofstarch solution Ras indicator.

Storage : protected from light.

0.01 M Ammonium and cerium nitrate. 3000200.

To 100.0 ml of0.1 M ammonium and cerium nitrateadd, with cooling, 30 ml of sulphuric acid R and dilute to1000.0 ml withwater R.

0.1 M Ammonium and cerium sulphate. 3000300.

Dissolve 65.0 g ofammonium and cerium sulphate R in amixture of 500 ml ofwater Rand 30 ml of sulphuric acid R.

Allow to cool and dilute to 1000.0 ml with water R.

Standardisation. To 25.0 ml of the ammonium and ceriumsulphate solution add 2.0 g ofpotassium iodide R and150 ml ofwater R. Titrate immediately with0.1 M sodiumthiosulphate, using 1 ml ofstarch solution Ras indicator.

0.01 M Ammonium and cerium sulphate. 3000400.

To 100.0 ml of0.1 M ammonium and cerium sulphateadd, with cooling, 30 ml of sulphuric acid R and dilute to1000.0 ml withwater R.

0.1 M Ammonium thiocyanate. 3000500.

Dissolve 7.612 g ofammonium thiocyanate Rin water Rand dilute to 1000.0 ml with the same solvent.

Standardisation. To 20.0 ml of0.1 M silver nitrate add25 ml ofwater R, 2 ml ofdilute nitric acid R and 2 mlofferric ammonium sulphate solution R2. Titrate withthe ammonium thiocyanate solution until a reddish-yellowcolour is obtained.



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Preparation of frequently used solutions

Content

1. Diluting Concentrated Acids (Last Login: 08/08/2009)2. Indicators(Last Login: 27/07/2009)3. Standard Buffer Solutions(Last Login: 27/07/2009)4. Special Solutions and Reagents (Last Login: 08/08/2009)

1. Diluting Concentrated Acids to 1 molar (1M) solutions1.1.General Safety Notes. Wear gloves and protect the eyes with safety goggles or

even better, a face shield. Dilution of concentrated acid should always be done in a

fume cupboard. Add concentrated acid to water slowly. Never add water to aconcentrated acid.

1.2.Hydrochloric acid:36% HCl1M HCl. Add 83.5 mL of 36% hydrochloric acid to about 600 mL ofdistilled water in a 1 litre measuring cylinder and make up to 1L.

32% HCl1M HCl. As above, except use 96 mL of 32% hydrochloric acid.

1.3.Nitric Acid:70% HNO31M HNO3. Add 62 mL of 70% nitric acid to about 700 mL of

distilled water in a 1 litre measuring cylinder and make up to 1 L.

1.4.Sulfuric acid:98% H2SO41M H2SO4. Add 54 mL of 98% concentrated sulfuric acid to

about 700 mL of distilled iced water in a 1 litre measuring cylinder and make up to1 L.

1.5.Acetic acid:99.5% CH3COOH (Glacial acetic acid)1M CH3COOH. Add 57 mL of theGlacial acetic acid to about 600 mL of distilled water in a 1 litre measuring

cylinder and make up to 1 L.


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2. IndicatorsIndicatorsaresubstances which change from one color to another when the hydrogen ionconcentration reaches a certain value, different for each indicator (1), and used to

determine the specified end-point in a chemical reaction or to indicate that a desiredchange in pH has been effected (2).

2.1. Litmus:Digest 25g of litmus powder with three successive, 100-mL portions of boiling

alcohol, continuing each extraction for about 1 hour. Filter, wash with alcohol,and discard the alcohol filtrate. Macerate the residue with about 25 mL of cold

water for 4 hours, filter, and discard the filtrate. Digest the residue with 125 mL ofboiling water for 1 hour, cool, and filter (2).

2.2. Methyl orange:Dissolve 1g of methyl orange in 1 liter of water. Filter if necessary (1).

2.3. Methyl red:Dissolve 100 mg of methyl red in 100mL of 95% ethyl alcohol. Filter ifnecessary (2).

2.4. Phenolphthalein:Dissolve 1g of phenolphthalein in 100mL of 95% ethyl alcohol (2).

Color in different pH conditionsIndicator

Acidic Neutral Basic (i.e. alkaline)Litmus Red

(pH < 5.0)Violet

(pH 5.0 - 8.0)Blue

(pH > 8.0)

Methyl orange Red(pH < 3.1)

Orange(pH 3.1 - 4.4)

Yellow(pH > 4.4)

Methyl red Red(pH < 4.2)

Orange(pH 4.2 - 6.3)

Yellow(pH > 6.3)

Phenolphthalein Colorless

(pH < 8.0)

Crimson pale

(pH 8.0 - 9.8)

Crimson

(pH > 9.8)


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3. Standard Buffer Solutions (According to USP 27)3.1.Components for Standard Buffer Solutions preparation

3.1.1. Potassium Biphthalate, 0.2 M. Dissolve 40.85g of Potassium Biphthalate[KHC6H4(COO)2] in water, and dilute with water to 1000 mL.

3.1.2. Potassium Phosphate, Monobasic, 0.2 M. Dissolve 27.22g of MonobasicPotassium Phosphate (KH2PO4) in water, and dilute with water to 1000 mL.

3.1.3. Boric Acid + Potassium Chloride, 0.2 M. Dissolve 12.37g of Boric Acid(H3BO3) and 14.91g of Potassium Chloride (KCl) in water, and dilute with

water to 1000 mL.3.1.4. Potassium Chloride, 0.2 M.Dissolve 14.91g ofPotassium Chloride (KCl) in

water, and dilute with water to 1000 mL.3.1.5. Carbon dioxide-free water is distillated water that has been boiled vigorously

for not less 5 minutes and allowed to cool without contact with atmosphere. 3.2.Composition of Standard Buffer Solutions (for volume 200 mL). Place 50mL of

solution 1 in a 200-mL volumetric flask, add the specified volume of the Solution 2

and add carbon dioxide-free water to volume 200 mL.

pH Solution 1, mL Solution 2, mL

Hydrochloric Acid Buffer

- 0.2 M KCl 0.2 M HCl

1.2 50 85.0

1.3 50 67.2

1.4 50 53.2

1.5 50 41.4

1.6 50 32.4

1.7 50 26.0

1.8 50 20.4

1.9 50 16.2

2.0 50 13.0

2.1 50 10.2

2.2 50 7.8

Acid Phthalate Buffer

- 0.2 M PotassiumBiphthalate

0.2 M HCl

2.2 50 49.5

2.4 50 42.2

2.6 50 35.42.8 50 28.9

3.0 50 22.3

3.2 50 15.7

3.4 50 10.4

3.6 50 6.3

3.8 50 2.9

4.0 50 0.1


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pH Solution 1, mL Solution 2, mL

Neutralized Phthalate Buffer

- 0.2 M Potassium

Biphthalate

0.2 M NaOH

4.2 50 3.0

4.4 50 6.64.6 50 11.1

4.8 50 16.5

5.0 50 22.6

5.2 50 28.8

5.4 50 34.1

5.6 50 38.8

5.8 50 42.3

PBS (Phosphate Buffer Solution)

- 0.2 M Monobasic

Potassium Phosphate

0.2 M NaOH

5.8 50 3.66.0 50 5.6

6.2 50 8.1

6.4 50 11.6

6.6 50 16.4

6.8 50 22.4

7.0 50 29.1

7.2 50 34.7

7.4 50 39.1

7.6 50 42.4

7.8 50 44.5

8.0 50 46.1

Alkaline Borate Buffer

- 0.2 M Boric Acid +Potassium Chloride

0.2 M NaOH

8.0 50 3.9

8.2 50 6.0

8.4 50 8.6

8.6 50 11.8

8.8 50 15.8

9.0 50 20.8

9.2 50 26.49.4 50 32.1

9.6 50 36.9

9.8 50 40.6

10.0 50 43.7


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4. Special Solutions and Reagents4.1. Bangs reagent (for glucose estimation). Dissolve 100 g of K2CO3, 66 g of

KCl and 160 g of KHCO3 in the order given in about 700 mL of water at 30C.Add 4.4 g of of CuSO4and dilute to 1 liter after the CO2is evolved. This

solution should be shaken only in such a manner as not allow entry of air. After24 hours 300 mL are diluted to 1 liter with saturated KCl solution, shaken

gently and used after 24 hours; 50 mL equivalent to 10 mg glucose (1)

4.2. Biuret Reagent. Dissolve 1.5 g of cupric sulfat and 6.0g of potassium sodiumtartrate in 500 mL of water in a 1000-mL volumetric flask. Add 300 mL ofcarbonate-free sodium hydroxide solution (1 in 10), dilute with carbonate-free

sodium hydroxide solution (1 in 10) to 1000mL and mix (2).

4.3. Bromine Water. Prepare a saturated solution of bromine by agitating 2-3 mL ofbromine with 100 mL of cold water in a glass-stoppered bottle (2).

4.4. Chlorine Water. Prepare a saturated solution of chlorine in water (2).4.5. Denigs Reagent. Mix 5 g of yellowmercuric oxide with 40 mL of water, and

while stirring slowly add 20 mL of sulfuric acid, then add another 40 mL ofwater, and stir until completely dissolved (2).

4.6. Lime water.Saturated solution of calcium hydroxide - 1.5 g of Ca (OH)2 in1000 mL of water . Use some excess, filter off CaCO3 and protect from CO2 ofthe air (1).

4.7. Mayers Reagent. Dissolve 1.358 g of mercuric chloride in 60 mL of water.Dissolve 5 g of potassium iodide in 10 mL of water. Mix the two solutions, and

dilute with water to 100 mL (2).

4.8. Nesslers Reagent. Dissolve 143 g of sodium hydroxide in 700 mL of water.Dissolve 50 g of red mercuric iodide and 40 g of potassium iodide in 200 mL ofwater. Pour the iodide solution into hydroxide solution, and dilute with water to

1000 mL. Allow the settle, and use the clear supernatant (2).4.9. Pasteurs salt solution. To 1000 mL of distilled water add 2.5 g of potassium

phosphate and o.25 g of calcium phosphate(1).4.10. Saline.Dissolve 9.0 g of sodium chloride in water to make 1000 mL (1).

References

1. CRC Handbook of Chemistry and Physics. 63rd

edition. Editor: R.C.Weast. CRC Press.

2. U.S. Pharmacopeia. USP 27 / NF 22.

Copyright 2009 by LABMANUAL Corporation. All rights reserved.


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BuffersA guide for the preparation and use of

buffers in biological systems

Advancing your life science discoveries


28/63

BuffersA guide for the preparation and use ofbuffers in biological systems

By

Chandra Mohan, Ph.D.

Copyright 2003 EMD Biosciences, Inc., An Affiliate of Merck KGaA, Darmstadt, Germany.All Rights Reserved.

A brand of EMD Biosciences, Inc.


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ii

A Word to Our Customers

We are pleased to present to you the newest edition of Buffers: A Guide for the

Preparation and Use of Buffers in Biological Systems. This practical resource has

been especially revamped for use by researchers in the biological sciences. This

publication is a part of our continuing commitment to provide useful product

information and exceptional service to you, our customers. You will find this

booklet a highly useful resource, whether you are just beginning your research

work or training the newest researchers in your laboratory.

Over the past several years, Calbiochem Biochemicals has clearly emerged as a

world leader in providing highly innovative products for your research needs in

Signal Transduction, including the areas of Cancer Biology, Alzheimers Disease,

Diabetes and Hypertension, Protein Kinase, G-Protein, Apoptosis, and Nitric

Oxide related phenomena. Please call us today for a free copy of our LATESTSignal Transduction Catalog and Technical Resource and/or our Apoptosis

Catalog.

If you have used Calbiochem products in the past, we thank you for your

support and confidence in our products, and if you are just beginning your

research career, please call us and give us the opportunity to demonstrate our

exceptional customer and technical service.

Please call us and ask for a current listing of our ever expanding Technical

Resource Library, now with over 60 Calbiochem publications. Or check out our

website at http://www.calbiochem.com for even more useful information.

Marie Bergstrom

Marketing Manager

CALBIOCHEM and Oncogene Research Products

CALBIOCHEM

A name synonymous with commitment to high quality and exceptional service.


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iii

Table of Contents:

Why Does Calbiochem Biochemicals Publish a Booklet on Buffers? . . . . . . . . . .1

Water, The Fluid of Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2

Ionization of Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3

Dissociation Constants of Weak Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . .4

Henderson-Hasselbach Equation: pH and pKa. . . . . . . . . . . . . . . . . . . . . . . . . . . . .5

Determination of pKa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6

pKaValues for Commonly Used Biological Buffers . . . . . . . . . . . . . . . . . . . . . . . . .7

Buffers, Buffer Capacity, and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8

Biological Buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10

Buffering in Cells and Tissues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10

Effect of Temperature on pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12

Effect of Buffers on Factors Other than pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13

Use of Water-Miscible Organic Solvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14

Solubility Equilibrium: Effect of pH on Solubility . . . . . . . . . . . . . . . . . . . . . . . .14

pH Measurements: Some Useful Tips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15

Choosing a Buffer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16

Preparation of Some Common Buffers for Use

in Biological Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18

Commonly Used Buffer Media in Biological Research . . . . . . . . . . . . . . . . . . . . .22

Isoelectric Point of Selected Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24

Isoelectric Point of Selected Plasma Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . .26

Approximate pH and Bicarbonate Concentration in

Extracellular Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26

Ionization Constants K and pKa for Selected Acids and

Bases in Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27

Physical Properties of Some Commonly Used Acids . . . . . . . . . . . . . . . . . . . . . . .27

Some Useful Tips for Calculation of Concentrations and

Spectrophotometric Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28

CALBIOCHEM Buffers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30


31/63

iv


32/63

1

Why Does Calbiochem Biochemicals Publish aBooklet on Buffers?

We are frequently asked questions on the use of buffers that we offer to researchlaboratories. This booklet is designed to help answer several basic questions

about the use of buffers in biological systems. The discussion presented here is

by no means complete, but we hope it will help in the understanding of general

principles involved in the use of buffers.

Almost all biological processes are pH dependent. Even a slight change in pH can

result in metabolic acidosis or alkalosis, resulting in severe metabolic complica-

tions. The purpose of a buffer in biological system is to maintain intracellularand extracellular pH within a very narrow range and resist changes in pH in the

presence of internal and external influences. Before we begin a discussion of

buffers and how they control hydrogen ion concentrations, a brief explanation

of the role of water and equilibrium constants of weak acids and bases is

necessary.


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2

Water: The Fluid of LifeWater constitutes about 70% of the mass of most living creatures. All biologi

reactions occur in an aqueous medium. All aspects of cell structure and funct

are adapted to the physical and chemical properties of water. Hence, it is

essential to understand some basic properties of water and its ionizationproducts, i.e., H+ and OH. Both H+ and OH influence the structure, assembly,

and properties of all macromolecules in the cell.

Water is a polar solvent that dissolves most charged molecules. Water dissolv

most salts by hydrating and stabilizing the cations and anions by weakening

their electrostatic interactions (Figure 1). Compounds that readily dissolve in

water are known as HYDROPHILIC compounds. Nonpolar compounds such a

chloroform and ether do not interact with water in any favorable manner anknown as HYDROPHOBIC compounds. These compounds interfere with

hydrogen bonding among water molecules.

Figure 1: Electrostatic interaction of Na+ and Cl ions and water molecules.

Several biological molecules, such as protein, certain vitamins, steroids, and

phospholipids contain both polar and nonpolar regions. They are known as

AMPHIPATHIC molecules. The hydrophilic region of these molecules are

arranged in a manner that permits maximum interaction with water molecule

However, the hydrophobic regions assemble together exposing only the small

area to water.


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Ionization of WaterWater molecules undergo reversible ionization to yield H+ and OH as per the

following equation.

H2O H+

+ OH

The degree of ionization of water at equilibrium is fairly small and is given b

the following equation where Keq is the equilibrium constant.

[H+][OH]Keq =

______________

[H2O]

At 25C, the concentration of pure water is 55.5 M (1000 18; M.W. 18.0).

Hence, we can rewrite the above equation as follows:

[H+][OH]Keq =

______________

55.5 M

or

(55.5)(Keq

) = [H+][OH]

For pure water electrical conductivity experiments give a Keqvalue of 1.8 x

10-16 M at 25C.

Hence, (55.5 M)(1.8 x 10-16 M) = [H+][OH]

or

99.9 x 10-16 M2 = [H+][OH]

or

1.0 x 10-14 M2 = [H+][OH]

[H+][OH], ion product of water, is always equal to 1.0 x 10-14 M2 at 25C. Wh

[H+] and [OH] are present in equal amounts then the solution gives a neutral

Here [H+][OH] = [H+]2

or

[H+] = 1 x 10-14 M2

and

[H+] = [OH] = 10-7 M

As the total concentration of H+ and OH is constant, an increase in one ion i

compensated by a decrease in the concentration of other ion. This forms the

basis for the pH scale.


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4

Dissociation Constants of Weak Acids and BasesStrong acids (hydrochloric acid, sulfuric acid, etc.) and bases (sodium hydroxide,

potassium hydroxide, etc.) are those that are completely ionized in dilute

aqueous solutions.

In biological systems one generally encounters only weak acids and bases. Weak

acids and bases do not completely dissociate in solution. They exist instead as an

equilibrium mixture of undissociated and dissociated species. For example, in

aqueous solution, acetic acid is an equilibrium mixture of acetate ion, hydrogen

ion, and undissociated acetic acid. The equilibrium between these species can be

expressed as:

k1

CH3COOH H+ + CH

3COO

k2

where k1 represents the rate constant of dissociation of acetic acid to acetate and

hydrogen ions, and k2 represents the rate constant for the association of acetate

and hydrogen ions to form acetic acid. The rate of dissociation of acetic acid,

-d[CH3COOH ]/dt, is dependent on the rate constant of dissociation (k1) and the

concentration of acetic acid [CH3COOH] and can be expressed as:

d [CH3COOH]

____________________ = k1

[CH3COOH]

dt

Similarly, the rate of association to form acetic acid, d[HAc]/dt, is dependent on

the rate constant of association (k2) and the concentration of acetate and

hydrogen ions and can be expressed as:

d [CH3COOH ]__________________ = k2 [H+] [CH3COO]dt

Since the rates of dissociation and reassociation are equal under equilibrium

conditions:k

1[CH

3COOH ] = k

2[H+] [CH

3COO]

ork1 [H

+] [CH3COO]_______ = ____________________k2 [CH3COOH]

and [H+] [CH3COO]Ka

= ___________________[CH

3COOH]

wherek

1_______ = Ka (Equilibrium constant)k2


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5

This equilibrium expression can now be rearranged to

[CH3COOH][H+] = Ka

_______________

[CH3COO]

where the hydrogen ion concentration is expressed in terms of the equilibrium

constant and the concentrations of undissociated acetic acid and acetate ion. The

equilibrium constant for ionization reactions is called the ionization constant or

dissociation constant.

Henderson-Hasselbach Equation: pH and pKaThe relationship between pH, pKa, and the buffering action of any weak acid and

its conjugate base is best explained by the Henderson-Hasselbach equation. In

biological experiments, [H+] varies from 10-1 M to about 10-10 M. S.P.L.

Sorenson, a Danish chemist, coined the p value of any quantity as the negative

logarithm of the hydrogen ion concentration. Hence, for [H+] one can write the

following equation:pH = log [H+]

Similarly pKa can be defined as log Ka. If the equilibrium expression is

converted to log then

[CH3COOH]

log [H+] = log Ka

log ______________[CH

3COO]

and pH and pKa substituted:

[CH3COOH]

pH = pKa

log ________________[CH3COO]

or[CH3COO]pH = pK

a+ log _______________

[CH3COOH]

When the concentration of acetate ions equals the concentration of acetic acid,

log [CH3COO]/[CH3COOH] approaches zero (the log of 1) and pH equals pKa (the

pKa of acetic acid is 4.745). Acetic acid and acetate ion form an effective

buffering system centered around pH 4.75. Generally, the pKa of a weak acid or

base indicates the pH of the center of the buffering region.

The terms pK and pKa are frequently used interchangeably in the literature. The

term pKa (a refers to acid) is used in circumstances where the system is being

considered as an acid and in which hydrogen ion concentration or pH is of


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6

interest. Sometimes the term pKb is used. pKb (b refers to base) is used when the

system is being considered as a base and the hydroxide ion concentration or pOH

is of greater interest.

Determination of pKapKa values are generally determined by titration. A carefully calibrated,

automated, recording titrator is used, the free acid of the material to be measured

is titrated with a suitable base, and the titration curve is recorded. The pH of the

solution is monitored as increasing quantities of base are added to the solution.

Figure 2 shows the titration curve for acetic acid. The point of inflection

indicates the pKavalue. Frequently, automatic titrators record the first derivative

of the titration curve, giving more accurate pKavalues.

Polybasic buffer systems can have more than one useful pKavalue. Figure 3

shows the titration curve for phosphoric acid, a tribasic acid. Note that the curve

has five points of inflection. Three indicate pKa1, pKa2 and pKa3, and two

additional points indicate where H2PO4 and HPO4

exist as the sole species.

Figure 2: Titration Curve for Acetic Acid

Figure 3: Titration Curve for Phosphoric Acid

pH

12

pKa1

= 2.12

pKa2= 7.21

pKa3

= 12.32

10

8

6

NaOH

4

2

pH

8

pKa= 4.766

4

2

NaOH0


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7

Table 1: pKaValues for Commonly Used Biological Buffers and Buffer Constituents

ADA, Sodium Salt 114801 212.2 6.60

2-Amino-2-methyl-1,3-propanediol 164548 105.1 8.83BES, ULTROL Grade 391334 213.2 7.15

Bicine, ULTROL Grade 391336 163.2 8.35

BIS-Tris, ULTROL Grade 391335 209.2 6.50

BIS-Tris Propane, ULTROL Grade 394111 282.4 6.80

Boric Acid, Molecular Biology Grade 203667 61.8 9.24

Cacodylic Acid 205541 214.0 6.27

CAPS, ULTROL Grade 239782 221.3 10.40

CHES, ULTROL Grade 239779 207.3 9.50

Citric Acid, Monohydrate, Molecular Biology Grade 231211 210.1 4.76

Glycine 3570 75.1 2.341

Glycine, Molecular Biology Grade 357002 75.1 2.341

Glycylglycine, Free Base 3630 132.1 8.40

HEPES, Free Acid, Molecular Biology Grade 391340 238.3 7.55

HEPES, Free Acid, ULTROL Grade 391338 238.3 7.55

HEPES, Free Acid Solution 375368 238.3 7.55

HEPES, Sodium Salt, ULTROL Grade 391333 260.3 7.55

HEPPS, ULTROL Grade 391339 252.3 8.00

Imidazole, ULTROL Grade 4015 68.1 7.00

MES, Free Acid, ULTROL Grade 475893 195.2 6.15

MES, Sodium Salt, ULTROL Grade 475894 217.2 6.15

MOPS, Free Acid, ULTROL Grade 475898 209.3 7.20MOPS, Sodium Salt, ULTROL Grade 475899 231.2 7.20

PIPES, Free Acid, Molecular Biology Grade 528133 302.4 6.80

PIPES, Free Acid, ULTROL Grade 528131 302.4 6.80

PIPES, Sodium Salt, ULTROL Grade 528132 325.3 6.80

PIPPS 528315 330.4 3.732

Potassium Phosphate, Dibasic, Trihydrate, Molecular Biology Grade 529567 228.2 7.213

Potassium Phosphate, Monobasic 529565 136.1 7.213

Potassium Phosphate, Monobasic, Molecular Biology Grade 529568 136.1 7.213

Sodium Phosphate, Dibasic 567550 142.0 7.213

Sodium Phosphate, Dibasic, Molecular Biology Grade 567547 142.0 7.213Sodium Phosphate, Monobasic 567545 120.0 7.213

Sodium Phosphate, Monobasic, Monohydrate, Molecular Biology Grade 567549 138.0 7.213

TAPS, ULTROL Grade 394675 243.2 8.40

TES, Free Acid, ULTROL Grade 39465 229.3 7.50

TES, Sodium Salt, ULTROL Grade 394651 251.2 7.50

Tricine, ULTROL Grade 39468 179.2 8.15

Triethanolamine, HCl 641752 185.7 7.66

Tris Base, Molecular Biology Grade 648310 121.1 8.30

Tris Base, ULTROL Grade 648311 121.1 8.30

Tris, HCl, Molecular Biology Grade 648317 157.6 8.30Tris, HCl, ULTROL Grade 648313 157.6 8.30

Trisodium Citrate, Dihydrate 567444 294.1

Trisodium Citrate, Dihydrate, Molecular Biology Grade 567446 294.1

1. pKa1 = 2.34; pKa2 = 9.60

2. pKa1 = 3.73; pKa2 = 7.96 (100 mM aqueous solution, 25C).

3. Phosphate buffers are normally prepared from a combination of the monobasic and dibasic salts, titrated againsteach other to the correct pH. Phosphoric acid has three pKa values: pKa1 = 2.12; pKa2 = 7.21; pKa3 = 12.32

Product Cat. No. M.W.pKa

at 20C


39/63

8

Buffers, Buffer Capacity and RangeBuffers are aqueous systems that resist changes in pH when small amounts of

acid or base are added. Buffer solutions are composed of a weak acid (the proton

donor) and its conjugate base (the proton acceptor). Buffering results from two

reversible reaction equilibria in a solution wherein the concentration of proton

donor and its conjugate proton acceptor are equal. For example, in a buffer

system when the concentration of acetic acid and acetate ions are equal, addition

of small amounts of acid or base do not have any detectable influence on the pH.

This point is commonly known as the isoelectric point. At this point there is no

net charge and pH at this point is equal to pKa.

[CH3COO]pH = pK

a+ log ________________

[CH3COOH]

At isoelectric point [CH3COO] = [CH3COOH] hence, pH = pKa

Buffer capacity is a term used to describe the ability of a given buffer to resist

changes in pH on addition of acid or base. A buffer capacity of 1 is when 1 mol

buffers solutions

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