equilibrium part 1 common ion effect. common ion effect whenever a weak electrolyte and a strong...

EQUILIBRIUMPart 1 Common Ion Effect

COMMON ION EFFECT

Whenever a weak electrolyte and a strong electrolyte share the same solution, the strong electrolyte dissociates completely and effects the solubility of the weak electrolyte.The weak electrolyte dissolves less than it would when it is by itself.

COMMON ION EFFECT

Keep in mind how to identify the strength of an acid.

The larger the Ka the stronger the acid. It means that the acid dissociates more into the ions.

Ka = [H+][A-] [HA]

EXAMPLE:

What is the pH of a solution of 0.30 mol acetic acid and 0.30 mol sodium acetate to added to enough water to make 1 L solution?

What should we start to create? Think of other equilibrium or acid-base problems.

Correct, a Table!

EXAMPLE:

Identify the strong and weak electrolytes. Identify the source of H+, so can find pH.

Sodium Acetate is a strong electrolyte, so need to find the Ka of acetic acid, Ka= 1.8 x10-5, and all H+ will come from scetic acid. Common ion is CH3COO-.

EXAMPLE:

[CH3COOH] [H+] [CH3COO-]

Initial [ ] 0.30 0 0.30

Change in [ ] -x +x +x

Equilibrium [ ] 0.30 – x x .30 +x

What are the initial concentrations (M) and the change in concentrations (variable)?

EXAMPLE:

[CH3COOH] [H+] [CH3COO-]

Initial [ ] 0.30 0 0.30



How can we use this information?

CALCULATING PH FROM KA

Ka = [H+][CH3COO-] [CH3COOH] Ka = (x) (0.30 + x) = 1.8 x 10-5

0.30 - xWe can assume that x is negligible compared to 0.30 since Ka is small.


Ka = (x) (0.30) = 1.8 x 10-5

0.30 x = 1.8 x 10-5

[H+] = 1.8 x 10-5 MpH = -log(1.8 x 10-5) = 4.74

This would have been pH = 2.64 if no common ion.

CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED

Calculate the fluoride ion concentration and pH of a solution that is 0.20 M HF and 0.10 M HCl.


Write the reactions in order to find the common ion. HF H+ + F-

HCl H+ + Cl-

The common ion is H+. HF is a weak acid so it does not completely dissociate, like HCl does.


What do we need to create?


A Table!How did we get this info?

[HF] [H+] [F-]

Initial [ ] 0.20 0.10 0


Equilibrium [ ] 0.20 – x 0.10 +x x


What do we need to find x?Where can we find Ka?

[HF] [H+] [F-]

Initial [ ] 0.20 0.10 0




Find Ka in Appendix D in Text.

[HF] [H+] [F-]

Initial [ ] 0.20 0.10 0




Ka = (.10 + x) (x) = 6.8 x 10-4

0.20 - x Simplifies if we assume x is relatively small compared to .10 or .20. Wy can we assume this?Ka = (.10 ) (x) = 6.8 x 10-4

.20


Ka = (.10 ) (x) = 6.8 x 10-4

.20

x = (.20 ) (6.8 x 10-4) = 1.4 x 10-3 = [F-] .10


Then we can use x to find the concentration (molarity) of H+ ions.

[H+] = .10 + .0014 = (roughly) .10

pH = 1.00

BUFFERED SOLUTIONS

Solutions that contain a weak conjugate acid-base pair are able to resist drastic changes in pH when small amounts of strong acids or strong bases are added are

called BUFFERS.

BUFFERS

Buffers have an acid to neutralize OH- and a base to neutralize H+. The acid and base must not consume each other. So the weak acid or weak base must be paired with a salt of the acid or base.

BUFFERS

Examples of buffers:CH3COOH and CH3COONa to get CH3COO-

OrNH4Cl and NH3 to get NH4

+

By choosing appropriate components and adjusting relative concentrations, a solution can be buffered at a pH.

BUFFERS

We need to follow some manipulation of formulas to see how this works.

HX is an acidMX is a salt of the acid where M is most likely an alkali metal

BUFFERS

HX is an acidMX is a salt of the acid where M is most likely an alkali metal SoHX (aq) H+ (aq) + X- (aq)The HX is the acid, H+ is from the acid dissociating, and the X- is from the acid and salt disscociating.

BUFFERS

HX (aq) H+ (aq) + X- (aq)The Ka isKa = [H+] [X-] [HX]So[H+] = Ka [HX] [X-]

BUFFERS

[H+] = Ka [HX] [X-][H+] is dependent on the Ka and the relationship of [HX] and [X-]

BUFFERS

[H+] = Ka [HX] [X-]If a base is added ([OH-]) then:OH-(aq) + HX(aq) H2O(l) + X-(aq)

And HX decreases and X- increases. Presence of HX counteracts the addition of base and pH increase is small.

BUFFERS

[H+] = Ka [HX] [X-]If an acid is added ([H+] or [H3O+]) then:H3O+(aq) + X-(aq) H2O(l) + HX(aq)

And HX increases and X- decreases. Presence of HX counteracts the addition of acid and pH decrease is small.

CALCULATING PH OF BUFFERS

[H+] = Ka [HX] [X-]

-log[H+] = -log -log[H+] = -logKa - log [HX] [X-]

CALCULATING PH OF BUFFERS

-log[H+] = -logKa - log [HX] [X-]

pH = pKa - log [HX] [X-]

pH = pKa + log [X-] [HX]

HENDERSON-HASSELBALCH EQUATION

pH = pKa + log [X-] [HX]

pH = pKa + log [base] [acid]Base and acid refer to the equilibrium concentration of the conjugate acid-base pair.

BUFFER CAPACITY

Amount of acid or base that a buffer can neutralize before the pH begins to change at a greater degree.

pH of a 1L solution of 1 M CH3COOH and 1 M CH3COONa has same pH as a 1L solution of .1 M CH3COOH and .1 M CH3COONa, just is a greater buffer.

PH RANGE

If the concentrations of weak acid and conjugate base pair is the same, then pH = pKa. So try to select a buffer based on pKa that is close to desired pH. Range that is good is pH = pKa + 1 .

STRONG ACID OR STRONG BASES AND BUFFERS

When a strong acid or base is added to a weak acid buffer, the strong acid or strong base is consumed. We need to calculate the new values of [HX] and [X-], then use that with Ka to calculate the [H+] and then pH.

CALCULATING PH OF BUFFER EXAMPLE:

What is the pH of a buffer that is 0.12 M lactic acid (HC3H5O3) and 0.10 M sodium lactate (NaC3H5O3)? Lactic Acid Ka = 1.4 x 10-4

What should we start to create? Think of other equilibrium or acid-base problems.

Correct, a Table!


Identify the strong and weak electrolytes. Identify the source of H+, so can find pH.

Sodium Lactate is a strong electrolyte, so need to find the Ka of lactic acid, Ka= 1.4 x10-4, and all H+ will come from lactic acid. Common ion is C3H5O3

-.


[HC3H5O3] [H+] [C3H5O3-]

Initial [ ] 0.12 0 0.10



What are the initial concentrations (M) and the change in concentrations (variable)?


How can we use this information?

[HC3H5O3] [H+] [C3H5O3-]

Initial [ ] 0.12 0 0.10




Ka = [H+][C3H5O3-]

[HC3H5O3] Ka = (x) (0.10 + x) = 1.4 x 10-4

0.12 - xWe can assume that x is negligible compared to 0.10 since Ka is small.


Ka = (x) (0.10) = 1.4 x 10-4

0.12 x = 1.8 x 10-4 (0.12 / 0.10)[H+] = 1.7 x 10-4 MpH = -log(1.7 x 10-4) = 3.77Or:pH = pKa + log [base] = 3.85 + (-0.08) [acid] still 3.77

EXAMPLE: PREPARING A BUFFER

How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer with a pH of 9.00? (assume the addition of NH4Cl does not change the volume of the solution)


Identify what we are dealing with. NH4Cl, NH4

+, Cl-, NH3, H2O

Cl- is a spectator ion.NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Kb = 1.8 x 10-5


NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Kb = 1.8 x 10-5 = [NH4+] [OH-]

[NH3]

pOH = 14.00- pH = 14.00 – 9.00 = 5.00

[OH-] = 1.0 x10-5


When we look at the Kb and notice that it is small and the common ion is already present, then we can use the initial concentration for NH3 of 0.10 M.


Then we can solve for [NH4]

Kb = 1.8 x 10-5 = [NH4+] [OH-]

[NH3]

[NH4] = 1.8 x10-5 [NH3] [OH-]


[NH4] = (1.8 x10-5 )( 0.10 M) (1.0 x10-5 M) = .18 MSo the answer is:(2.0 L) .18 mol NH4 =0.36 mol NH4Cl L

EXAMPLE: CALCULATING PH CHANGES IN BUFFERS

A buffer that is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.000 L of solution has a pH of 4.74. (1) calculate the pH of the solution after 5.0 mL of 4.0 M NaOH (aq) is added. (2) Then compare the pH of a solution of adding 5.0 mL of 4.0 M NaOH (aq) added to 1.000 L H2O.


What should we make? A variation of the table for the reaction:CH3COOH + OH- CH3COO- + H2O(l)

CH3COOH OH- CH3COO-


CH3COOH + OH- CH3COO- + H2O(l)Water does not show up because of species. We also focus on mol not [ ]

CH3COOH OH- CH3COO-

Buffer before 0.300 mol

0 0.300 mol

Addition .020 mol

Buffer after 0.280 mol 0 .320 mol


Now look at the equilibrium of: CH3COOH CH3COO- + H+ And we can use the info in the table to help us find the concentrations (molarity) keeping in mind the volume has changed due to addition of the base


The concentrations are [CH3COOH] = 0.280 mol / 1.005 L[CH3COO- ] = 0.320 mol / 1.005 L

[CH3COOH] = 0.279 M[CH3COO- ] = 0.318 M


With these concentrations: [CH3COOH] = 0.279 M[CH3COO- ] = 0.318 MUse Henderson-Hasselbalch Equation:pH = pKa + log [X-] [HX]


Use Henderson-Hasselbalch Equation:pH = 4.74 + log .318 M = 4.80 .279 M

Remember that the pKa is the same as the pH when the acid and conjugate base concentrations are equal.


Now if we compare that change in pH to the change in pH with the same amount of base added to just water, we will see what a difference a buffer makes.


We would still have the OH- added to the water. The concentration would then be 0.20 mol OH- / 1.005 L = 0.020 M OH- Then the pH can be found by:pH = 14.00 – pOHpH= 14.00 – (-log 0.020 )pH = 14.00 - 1.70pH = 12.3


In the first example with the buffer, the pH increased by .06

Without the buffer, the pH increased from 7.00 to 12.3 (increase of 5.30)

equilibrium part 1 common ion effect. common ion effect whenever a weak electrolyte and a strong...

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