pH, Ka, pKa, KwpH of strong and weak acid and strong

baseAcid-base indicators

Change in pH of acid-base titrationBuffer solution

Solubility product, Ksp

Common ion effect

Ionic Equilibria

The pH scale pH is a measure of [H+(aq)] in an aqueous

solution. pH = -log10[H+] In most aqueous solution, the pH is between

the range of 0 – 14. pH values depend on concentration of acid or

base and degree of dissociation,α. pH of an acid increases upon dilution at

constant temperature to a volume V.

pH of a base decreases upon dilution at constant temperature to a volume V.

Therefore, using the dissociation constant to measure pH is more reliable because it is a constant at all dilution and only will be influenced by temperature.

Ionic product of waterWater is slightly ionised :H2O H+ + OH- H = positiveKc = [H+] . [OH-]

[H2O]Degree of ionisation of water is very small, hence [H2O]

remains virtually constant.[H+].[OH-] = Kc x [H2O] = constant

= Kw

Kw = [H+] . [OH-]

Kw is the ionic product of water.Kw depend on temperature.At 25C, Kw = 1.0 x 10-14 mol2dm-6. In pure water, 25C : [H+] = [OH-]Hence [H+] = [OH-] = (1.0 x 10-14)1/2

= 1.0 x 10-7 mol dm-3

Therefore, pH of water = -log10(1.0 x 10-7 ) = 7.0

The relationship between pH, [H+] and [OH-] :

Kw varies with temperature. When temperature increases, Kw increase.

Will the pH of water be affected when temperature is changed?

E.g : Calculate the pH of water at the temperature below, using the Kw values given.

a)At 10C, Kw = 0.1 x 10-14 mol2 dm-6.

b)At 100C, Kw = 51 x 10-14 mol2 dm-6.

Sometimes the term pOH is used.pOH = -log10[OH-]At 25C, pH + pOH = 14

pOH = 14 - pH

Calculate the pH of strong acid and baseExample: Calculate the pH of the following

solutions.1.0.1 mol dm-3 HCl

2.0.001 mol dm-3 H2SO4

3.A solution where [H+] ions = 3.5 x 10-3 mol dm-3

The pH of a solution of HCl is 3.30. Calculate the concentration of hydrogen ions.

Calculate the pH of strong acid and base.

1.

2.

3.

4.

pH of Weak Acids and BasesAcid Dissociation Constant, Ka

Weak Acid dissociates partially in water :HA(aq) + H2O(l) H3O+(aq) + A-(aq)

The acid dissociation constant, Ka :

Ka = [H3O+] [A-] mol dm-3

[HA]pKa = -log10 Ka

The Ka value is the measure of the strength of acids.

Larger Ka value (smaller pKa), stronger acid.

Base Dissociation Constant, Kb

Weak base dissociates partially in water :B(aq) + H2O(l) BH+(aq) + OH-(aq)

The base dissociation constant, Kb :

Ka = [BH+] [OH-] mol dm-3

[B]pKb = -log10 Kb

The Kb value is the measure of the strength of base.

Larger Kb value (smaller pKb), stronger base.

Calculating the pH of a weak acidE.g 1:What is the pH of 0.100 mol dm-3 ethanoic

acid ? Ka = 1.74 x 10-5 mol dm-3.

E.g 2 :What is the pH of 0.05 mol dm-3 methanoic

acid if its pKa is 3.75?

E.g 3 :The pH of a weak acid, HA, of concentration

0.1 mol dm-3 was found to be 4.00. Calculate the value of pKa for the acid.

Exercise :1.Ka for a weak monobasic acid = 1.0 x 10-5 mol dm-

3. In a 0.1 mol dm-3 solution of the acid, calculatea) the concentration of H+(aq) ions;b) the pH;c) the concentration of OH-(aq) ions.

[Kw = 1.0 x 10-14 mol2 dm-6]

2. A 0.1 mol dm-3 solution of a weak monobasic acid has a pH of 4.0. Calculate Ka and pKa for the acid.

3. The acid dissociation constant, Ka, for methanoic acid is 1.8 x 10-4 mol dm-3. In a 0.1 mol dm-3 solution of methanoic acid, calculate:a) the concentration of hydrogen ions;b) the pH;c) the concentration of hydroxide ions;d) the degree of ionisation of then acid.

Calculating the pH of a weak base

E.g 1 :What is the pH of 0.100 mol dm-3 NH3(aq) if Ka for

the NH4+ ion is 5.62 x 10-10 mol dm-3? Kw = 1.00 x

10-14 mol2 dm-6.

E.g 2 :Sodium ethanoate solution is alkaline because the ethanoate ion is a weak base and reacts with water according to the equation :CH3CHOO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

a) pKa for ethanoic acid is 4.76. Calculate a value for Ka, including the units.

b) Write an expression for Ka for ethanoic acid, and rearrange it to give an expression for [H+].

c) Kw is 1.00 x 10-14 mol2 dm-6. Write an expression for Kw and rearrange it to give an expression for [OH-]

d) Write an expression for the concentration of CH3COOH at equilibrium.

e) What is the equilibrium concentration of CH3COO- ions in 0.2 mol dm-3 sodium ethanoate solution? State any assumption you are making.

f) By combining your answers to (b), (d), and (e), calculate the pH of 0.2 mol dm-3 sodium ethanoate.

Exercise : Calculate the pHs of the following solutions of

weak bases. In each case. Kw = 1.00 x 10-14.

(a) 0.05 mol dm-3 1-aminopropane solution, C3H7NH2.

pKa for C3H7NH3+ = 10.84

C3H7NH2 + H2O C3H7NH3+ + OH-

(b) 0.005 mol dm-3 phenylamine solution, C6H5NH2.

pKa for C6H5NH3+ = 4.62

C6H5NH2 + H2O C6H5NH3+ + OH-

Buffer Solution*Is a solution whose pH remains almost

unchanged when small amounts of acid or base are added to it.

2 types of buffer solution:1) Acidic buffer mixture of weak acid with sodium salt of weak acid. E.g, CH3COOH and CH3COO-Na+.

2) Alkaline buffer mixture of a weak base with the salt of the weak base. E.g, NH3 and NH4Cl.

Buffer works by removing extra acid or alkali added.

Acidic BufferCH3COOH CH3COO- + H+

CH3COO-Na+ CH3COO- + Na+ (Salt is fully ionised, high [CH3COO-])

Weak acid will only be slightly dissociated hence concentration of un-ionised CH3COOH is relatively high.

When small amount of acid added to buffer:H+ + CH3COO- CH3COOH

(added) (from salt)

The H+ ions added combine with CH3COO- ion from the salt to form CH3COOH (additional H+ ions removed). Hence pH remains almost unchanged.

[salt] in buffer decrease & [acid] increase is equal to the amount of H+ added.

• When small amount of base is added to buffer : OH- + CH3COOH CH3COO- + H2O

(added)

The extra OH- ions added removed by reacting with the large concentration of CH3COOH. Hence pH remains almost unchanged.

[acid] in buffer decrease and [salt] increase is equal to the amount of OH- added.

Alkaline BuffersNH3 + H2O NH4

+ + OH- (weak base partially ionised, concentration of NH3 is relatively high)

NH4Cl NH4+ + Cl- (salt is fully ionised)

When small amount of acid added to buffer:H+ + NH3 NH4

+

(added)

The extra H+ ions added removed by reacting with the large concentration of NH3 in the buffer. Hence pH remains almost unchanged.

[base] decrease and [salt increase is equal to the amount of H+ added.

• When small amount of base is added to buffer : OH- + NH4

+ NH3 + H2O

(added) (from salt)

The OH- ions added combine with NH4+ ion from

the salt to form NH3 (additional OH- ions removed). Hence pH remains almost unchanged.

[salt] decrease and [NH3] increase is equal to the amount of OH- added.

Application of Buffer Solution1) HCO3

- ion act as the principle buffer in human blood.

When acid enters the blood, the extra H3O+ ions are removed by HCO3

- ions which act as a base:

HCO3- + H3O+ H2CO3 + H2O

Or HCO3- + H+ H2CO3

Or HCO3- + H+ H2O + CO2

When base enters the blood, extra OH- ion are removed by carbonic acid.

H2CO3 + 2OH- CO3 2- + 2H2O

2) H2PO4- and HPO4

2- ions also act as a buffer in blood or saliva.

H2PO4- ion is a weak acid and dissociates partially in aqueous solution :

H2PO4- HPO4

2- + H+

When acid is added, the equilibrium shift to the left, extra H+ is consumed, pH maintained. [H2PO4

-] increases.

HPO42- + H+ H2PO4

- When alkali is added, equilibrium shift to the right

because the OH- ion added will react with the H+ ions in the buffer.

OH- + H+ H2O

H2PO4- + OH- HPO4

2- + H2O OH- ions removed, pH maintained.

3) Other usage of buffer solution : Solution of known pH for checking

instruments and indicators. Biochemistry experiments.

Calculation of pH of buffer solutionIn an acidic buffer :

HA H+ + A- (partial dissociation)MA M+ + A-

Ka = [H+] . [A-]

[HA] Where [HA] = concentration of HA acid in the mixture.

[A-] = concentration of A- from MA (the amount of A- from HA is very small, can be ignored)[H+] = Ka x [HA]

[A-]

[H+] ≈ Ka x [acid] or pH = pKa + log [salt]

[salt] [acid]This equation explains why the pH is affected very

little by dilution because the ratio of [acid]/[salt] remains constant on dilution.

[H+] ≈ Ka

Example 1 : Acidic bufferA buffer solution made by mixing 3.28 g dm-3 of CH3COO-

Na+ and 0.01 mol dm-3 CH3COOH. [Ka (CH3COOH) = 1.7 x 10-5 mol dm-3]

(a)What is the pH of the buffer.

(b) Calculate the change in pH when 1 cm3 of 1.0 mol dm-3 NaOH is added to 1 dm3 of buffer.

(c) Calculate the change in pH when 1 cm3 of 1.0 mol dm-3 NaOH is added to 1 dm3 of 0.01 mol dm-3 CH3COOH.

Example 2 : Alkaline buffer. A buffer solution contains 1.00 mol dm-3 NH3 and 0.40 mol dm-

3 NH4Cl. [Ka(NH4+) = 5.62 x 10-10 mol dm-3]

(a) Calculate the pH of the buffer.

(b) Calculate the effect on the pH when 5.00 cm3 of 10.0 mol dm-3 HCl solution is added to 1000cm3 of the buffer solution.

Exercise:

In what proportions should ammonia and ammonium chloride be mixed in solution to give a buffer solution of pH 10.0? pKa (NH4

+) = 9.25.

Hydrolysis of Salts

Salt of Solution formed in water

Reaction (example)

Weak acid & strong base

Alkaline solution

CH3COONa(aq) CH3COO-(aq) + Na+(aq)-CH3COO- ion reacts with H+ ions from water.CH3COO- + H+(aq) CH3COOH(aq)-Most H+ removed because CH3COOH is a weak acid. [H+] decrease, pH > 7.- [OH-] > [H+]

Strong acid & weak base

Acidic solution (NH4)2SO4(aq) 2NH4+(aq) + SO4

2-(aq)- NH4

+ ions react with OH- ions from water.NH4

+(aq) + OH- (aq) NH3 (aq) + H2O(l)-Most OH- removed because NH3 is a weak base.-As OH- ions are removed, more water ionises to keep Kw constant. [H+] increases, pH < 7- [H+] > [OH-]

In water, [H+] = [OH-] Neutral

Hydrolysis of Salts

Salt of Solution formed in water

Reaction (example)

Weak acid & weak base

Neutral solution

CH3COONH4(aq) CH3COO- (aq) + NH4+

(aq)-CH3COO- ions react with H+ from water to form CH3COOH. NH4

+ ion react with OH- ions from water to form NH3 & H2O.-Both H+ and OH- are equally removed, pH remains at 7.- [H+] = [OH-]

Strong acid & strong base

Neutral solution

NaCl Na+ + Cl-

-NaCl just dissolves in water hence concentration of H+ & OH- remains. [H+] = [OH-]

Acid-base IndicatorsAcid-base indicators are substances which change

colour according to the hydrogen ions concentration of the solution to which they are added.

Most indicators are weak acids. They have an acid colour (unionised molecule, HIn) and the alkaline colour (anion, In-).

Each indicator has a pH range over which it changes colour.

Example:

How an indicator works.Indicators can be regarded as weak acids.HIn(aq) H+(aq) + In-(aq)

Ka = KIn = [H+] . [In-]

[HIn]At the end point of titration, [HIn] = [In-] KIn = [H+]

pKIn = pHAt the end point, both acid colour and alkaline colour

form will be present in appreciable quantities ([HIn] = [In-])

It is not possible to determine precisely when the two forms are at equal concentration. Hence indicators change colour over a range of about 2 pH units.

In titration, an acid-base indicator is used to mark the end-point. For an accurate result :

a)The indicator must change colour sharply on additon of 1 or 2 drops of liquid from the burette.

b)The colour change must occur when the correct volume of liquid is added from the burette.

pH change during titrations1. Strong acid + strong base titration.50cm3 of 0.1 mol dm-3 NaOH + 25 cm3 of 0.1 mol

dm-3 HCl.

pH curve begins at pH 1, indicating the presence of strong acid.

pH curve ends at pH 13 , indicating presence of strong base.

As NaOH is added, the pH increases slowly, but increases rapidly between the range of 3 – 11.

The sharp increase in pH indicates the end point of titration.

The end-point is the mid-point of the vertical line of the titration curve (pH = 7).

Almost any indicator can be used but bromothymol blue is the ideal indicator.

2. Strong acid + weak base50cm3 of 0.1 mol dm-3 NH3 + 25 cm3 of 0.1 mol

dm-3 HCl.

pH curve begins at pH 1, indicating the presence of strong acid.

The graph ends at pH 11, showing presence of weak base.

As NH3 is added, there’s a slow increase in pH, but increases rapidly between the range of 3 – 7.

The end-point of the solution is acidic (pH ≈ 5.5).This is due to hydrolysis of NH4Cl salt.The vertical section of the curve is shorter than

titration of strong acid and strong base.Methyl orange is the ideal indicator.Phenolphthalein cannot be used because it will

change colour at the wrong volume of NH3 solution (about 30 cm3).

3. Weak acid + strong base50cm3 of 0.1 mol dm-3 NaOH + 25 cm3 of 0.1 mol

dm-3 CH3COOH.

pH curve begins at pH≈3, indicating the presence of weak acid.

The graph ends at pH≈13, showing presence of strong base.

When half the acid is neutralised, [CH3COOH] = [CH3COONa] max. buffer capacity

This is a buffer solution and the pH only changes gradually at the buffer region.

The end-point of the solution is alkaline (pH ≈ 8.5).This is due to hydrolysis of CH3COONa salt.The vertical section of the curve is shorter than

titration of strong acid and strong base.Phenolphthalein is the ideal indicator.Methyl orange is unsuitable because it will change

colour very slowly over a large volume of NaOH.

The buffer region corresponds to pH = pKa ± 1.

When : [acid] = 10 , pH = pKa -1

[salt] 1

[acid] = 1 , pH = pKa

[salt] 1

[acid] = 1 , pH = pKa +1

[salt] 10Acidic buffer is most effective (at max. buffer

capacity) when [acid]=[salt], pH = pKa

Alkaline buffer is most effective (at max. buffer capacity) when [base]=[salt], pOH = pKb

In the buffer region, the pH is insensitive to small changes in concentration of acid or base.

4. Weak acid + weak base50cm3 of 0.1 mol dm-3 NH3 + 25 cm3 of 0.1 mol

dm-3 CH3COOH.

pH curve begins at pH≈3, indicating the presence of weak acid.

The graph ends at pH≈11, showing presence of weak base.

At the end point, the solution (CH3COONH4) is approximately neutral.

Although the salt is hydrolysed, both H+ and OH- from water are removed equally. No net change in pH.

There is no straight vertical section on the graph.No indicator can be used to detect the end-point as

the colour change is always gradual.A pH meter is used during titration

potentiaometric titration.To find the end-point, a graph of pH vs. volume must

be drawn, and the end-point (at pH 7) determined from the graph.

How to find end point pH and volume when the graph has no vertical section?

5. Titration of polybasic acidsE.g 1 : 0.1 mol dm-3 NaOH gradually added to

20 cm3 of 0.1 mol dm-3 H2SO4.

H2SO4 is a dibasic acid. It reacts with NaOH in 2 steps. Hence it has 2 end-points.H2SO4 + NaOH NaHSO4 + H2O

NaHSO4 + NaOH Na2SO4 + H2O

The 1st end-point detected after 20 cm3 of NaOH added can be detected using methyl orange.

A solution of NaHSO4 is left in the titration flask.The 2nd end-point detected after 40 cm3 of NaOH

added can be detected using phenolphthalein.A solution of NaSO4 is formed.

E.g 2 : 0.1 mol dm-3 NaOH gradually added to 20 cm3 of 0.1 mol dm-3 H3PO4.

Phosphoric acid is tribasic.It reacts with NaOH in 3 steps. Hence 3 end-points

are detected. H3PO4 + NaOH NaH2PO4 + H2O

NaH2PO4 + NaOH Na2HPO4 + H2O

Na2HPO4 + NaOH Na3PO4 + H2O

Solubility product When an ionic compound, AnBm, is slightly soluble in water,

some solid dissolves to form a saturated solution. AnBm(s) nA+(aq) + mB-(aq) When the solution is saturated, the mixture is at

equilibrium at a given temperature.

Ksp = [A+]n.[B-]m

Ksp is called solubility product of the ionic solid. Solubility product only apply to slightly soluble ionic

compounds. The Ksp expression shows the interaction between ions in

the solution.

More soluble salts, higher Ksp value

Less soluble salts, lower Ksp value.

Ksp value are only affected by temperature.

Temperature increase, Ksp value increase.

Calculating Ksp from concentrations.Example 1 : The solubility of calcium

sulphate, CaSO4 at 298K is 0.67 g dm-3. Calculate the solubility product at this temperature.

Example 2 : The solubility of lead (II) chloride, PbCl2 is 0.016 mol dm-3 at 298K. Calculate the solubility product at this temperature.

Exercise

6.

Calculating solubility from Ksp

Example 1 : the solubility product of calcium carbonate, CaCO3 = 5.0 x 10-9 mol2 dm-6. Calculate the solubility of CaCO3 in g dm-3

Example 2 :Calculate the solubility of silver chloride, AgCl at 298K if the solubility product is 1.8 x 10-10 mol2 dm-6.

Example 3 : Calculate the solubility in g dm-3 of chromium (III) hydroxide, Cr(OH)3, at 25C if its solubility product is 1.0 x 10-33 mol4 dm-12.

Exercise 1.

2.

3.

4.

5.

Common ion effectThe solubility of an ionic compound of a solution is

decreased if the solution already contains one of the ions.

Example 1 : A saturated solution of calcium sulphate.CaSO4(s) Ca2+(aq) + SO4

2-(aq)

Ksp = [Ca2+][SO42-] = 2.4 x 10-5 mol2 dm-6 at 298K

When a solution that contains the same ion is added to the saturated CaSO4 solution, for e.g. dilute H2SO4 equilibrium shift to left, CaSO4 will precipitate out (or solubility decrease).

In presence of the additional “common ion” (SO42-) the

CaSO4 becomes less soluble.If a solution that contains Ca2+ ions is used, it will have

the same effect. CaSO4 will precipitate out.

Example 2 : Solubility of AgCl in dilute HCl is less than in pure water because of the “common ion”, Cl- ion from the HCl.

AgCl(s) Ag+(aq) + Cl-(aq)HCl(aq) H+(aq) + Cl-(aq)The presence of Cl- ion from HCl cause the

equilibrium to shift left decrease in solubility of AgCl.

Calculation on the common ion effect.E.g 1 : Ksp for AgCl is 2.0 x 10-10 mol2 dm-6.

Calculate the solubility of AgCl in(a) water(b) 0.1 mol dm-3 HCl

E.g 2 : Calculate the solubility of calcium sulphate in mol

dm-3 in (a) water(b) 0.50 mol dm-3 dilute H2SO4

Ksp = 2.4 x 10-5 mol2 dm-6

Uses of Ksp

1) Predicting precipitation using Ksp

Ksp can be used to predict whether precipitates will form when solutions are mixed.

For a sparingly soluble salt, AxBy :

AxBy(s) xAy+(aq) + yBx-(aq)

If [Ay+]x [Bx-]y = Ksp solution is saturated.

If [Ay+]x [Bx-]y < Ksp solution is not saturated, no ppt formed.

If [Ay+]x [Bx-]y >Ksp ppt forms.

[Ay+]x [Bx-]y is called ionic product.

E.g : Will a precipitate of PbCl2 be formed if 10 cm3 of 0.10 mol dm-3 Pb(NO3)2 is mixed with 10 cm3 of 0.20 mol dm-3 HCl? Ksp(PbCl2) = 1.6 x 10-5 mol3 dm-9 at 298K.

Exercise 1.Will a precipitate of Ca(OH)2 form if 5.0 cm3 of

0.050 mol dm-3 NaOH solution is added to 5.0 cm3 of 0.050 mol dm-3 CaCl2 solution? Ksp(Ca(OH)2) = 5.5 x 10-6 mol3 dm-9 at 298K.

2. Will a precipitate of Ca(OH)2 form if 5.0 cm3 of NH3 solution containing 2.0 x 10-3 mol dm-3 of OH- ions is added to 5.0 cm3 of 0.050 mol dm-3 CaCl2 solution? Ksp(Ca(OH)2) = 5.5 x 10-6 mol3 dm-9 at 298K.