# Ionic Equilibria pH, Ka, pKa, Kw

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Ionic Equilibria pH, Ka, pKa, Kw
pH of strong and weak acid and strong base Acid-base indicators Change in pH of acid-base titration Buffer solution Solubility product, Ksp Common ion effect

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 water Water is slightly ionised :
H2O H+ + OH- H = positive Kc = [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 25C, Kw = 1.0 x mol2dm-6. In pure water, 25C : [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. At 10C, Kw = 0.1 x mol2 dm-6. At 100C, Kw = 51 x mol2 dm-6.

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

Calculate the pH of strong acid and base
Example: Calculate the pH of the following solutions. 0.1 mol dm-3 HCl 0.001 mol dm-3 H2SO4 A solution where [H+] ions = 3.5 x 10-3 mol dm-3 The pH of a solution of HCl is Calculate the concentration of hydrogen ions.

Calculate the pH of strong acid and base.
1. 2.

3. 4.

pH of Weak Acids and Bases
Acid 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 acid
E.g 1: What is the pH of mol dm-3 ethanoic acid ? Ka = 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 Calculate the value of pKa for the acid.

Exercise : 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, calculate a) the concentration of H+(aq) ions; b) the pH; c) the concentration of OH-(aq) ions. [Kw = 1.0 x mol2 dm-6]

A 0. 1 mol dm-3 solution of a weak monobasic acid has a pH of 4
A 0.1 mol dm-3 solution of a weak monobasic acid has a pH of Calculate Ka and pKa for the acid.

The acid dissociation constant, Ka, for methanoic acid is 1
The acid dissociation constant, Ka, for methanoic acid is 1.8 x 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 mol dm-3 NH3(aq) if Ka for the NH4+ ion is x mol dm-3? Kw = 1.00 x 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) pKa for ethanoic acid is Calculate a value for Ka, including the units. Write an expression for Ka for ethanoic acid, and rearrange it to give an expression for [H+]. Kw is 1.00 x mol2 dm-6. Write an expression for Kw and rearrange it to give an expression for [OH-] Write an expression for the concentration of CH3COOH at equilibrium. What is the equilibrium concentration of CH3COO- ions in 0.2 mol dm-3 sodium ethanoate solution? State any assumption you are making. 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 (a) 0.05 mol dm-3 1-aminopropane solution, C3H7NH2. pKa for C3H7NH3+ = 10.84 C3H7NH2 + H2O C3H7NH OH- (b) mol dm-3 phenylamine solution, C6H5NH2. pKa for C6H5NH3+ = 4.62 C6H5NH2 + H2O C6H5NH 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 Buffer CH3COOH 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 Buffers NH3 + H2O NH OH- (weak base partially ionised, concentration of NH3 is relatively high) NH4Cl  NH Cl- (salt is fully ionised) When small amount of acid added to buffer: H NH 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 Solution
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: HCO H3O H2CO3 + H2O Or HCO H+  H2CO3 Or HCO H+  H2O + CO2 When base enters the blood, extra OH- ion are removed by carbonic acid. H2CO OH-  CO H2O

H2PO4- and HPO42- ions also act as a buffer in blood or saliva.
H2PO4- ion is a weak acid and dissociates partially in aqueous solution : H2PO HPO H+ When acid is added, the equilibrium shift to the left, extra H+ is consumed, pH maintained. [H2PO4-] increases. HPO 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 H2PO OH-  HPO H2O OH- ions removed, pH maintained.

Other usage of buffer solution :
Solution of known pH for checking instruments and indicators. Biochemistry experiments.

Calculation of pH of buffer solution
In 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 buffer A 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
(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+) = x 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 In water, [H+] = [OH-] Neutral 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) + SO42-(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-]

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 NaCl  Na Cl- -NaCl just dissolves in water hence concentration of H+ & OH- remains.

Acid-base Indicators Acid-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 = pH At 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
In titration, an acid-base indicator is used to mark the end-point. For an accurate result : The indicator must change colour sharply on additon of 1 or 2 drops of liquid from the burette. The colour change must occur when the correct volume of liquid is added from the burette.

pH change during titrations
1. 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 base 50cm3 of 0.1 mol dm-3 NH 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 base 50cm3 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] [acid] = 1 , pH = pKa [salt] 1 [acid] = 1 , pH = pKa +1 [salt] Acidic buffer is most effective (at max. buffer capacity) when [acid]=[salt],  pH = pKa

Alkaline buffer is most effective (at max
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 base 50cm3 of 0.1 mol dm-3 NH 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 acids
E.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
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
Example 2 : The solubility of lead (II) chloride, PbCl2 is 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 mol2 dm-6.

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

Exercise 1.

2. 3. 4. 5.

Common ion effect The 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) + SO42-(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 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 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
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 Will a precipitate of Ca(OH)2 form if 5.0 cm3 of mol dm-3 NaOH solution is added to 5.0 cm3 of 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
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 mol dm-3 CaCl2 solution? Ksp(Ca(OH)2) = 5.5 x 10-6 mol3 dm-9 at 298K.