Ionic Equilibria Notes

Salts are formed when acids react with bases in a neutralization reaction.
There are three different types of salts:
- Neutral salt (SA-SB): formed from strong acid and strong base.
- Acidic salt (SA-WB): formed from strong acid and weak base.
- Basic salt (SB-WA): formed from strong base and weak acid.

Hydrolysis is the reaction of a cation or an anion (or both) with water.
The reaction is reversible.
Example:
- $X^+ (aq) + H_2O (l) \rightleftharpoons$
- $Y^- (aq) + H_2O (l) \rightleftharpoons$

The relationship between the dissociation constants of acids ( $Ka$ ) and their conjugate base ( $Kb$ ) can be derived as follows:
- $CH_3COO^-$ hydrolysis in water:
 - Acid: $CH_3COOH$
 - Conjugate Base: $CH_3COO^-$
 - Base: $CH_3COO^-$
 - Conjugate Acid: $CH_3COOH$
$K_w = [H^+][OH^-]$
$Ka = \frac{[CH3COO^-][H^+]}{[CH_3COOH]}$
$Kb = \frac{[CH3COOH][OH^-]}{[CH_3COO^-]}$
$Ka \times Kb = \frac{[CH3COO^-][H^+]}{[CH3COOH]} \times \frac{[CH3COOH][OH^-]}{[CH3COO^-]} = [H^+][OH^-] = K_w$
- Therefore, $Ka \times Kb = K_w$

Neutral salts are formed from the neutralization reaction of strong acids (SA) with strong bases (SB).
The pH of aqueous solutions of neutral salts is 7 ( $pH = 7$ ).
Examples: $NaCl$ , $KBr$ , $KI$ , $RbBr$ , $BaCl_2$
$HCl (aq) + NaOH (aq) \rightarrow NaCl (aq) + H_2O (l)$
- Strong Acid + Strong Base -> Neutral Salt
In aqueous solution, $NaCl$ dissociates completely to $Na^+$ and $Cl^-$ ions.
Ions of neutral salt do not undergo hydrolysis
- $Na^+ + H_2O \rightarrow No Reaction$
- $Cl^- + H_2O \rightarrow No Reaction$

Acidic salts are formed when strong acids (SA) react with weak bases (WB).
The pH of aqueous solutions of acidic salts is less than 7 (pH < 7).
The cation of acidic salts will undergo hydrolysis.
Examples: $NH4Cl$ , $NH4NO_3$
$HCl (aq) + NH3 (aq) \rightarrow NH4Cl (aq)$
- Strong Acid + Weak Base -> Acidic Salt
In aqueous solution, $NH4Cl$ dissociates completely to $NH4^+$ and $Cl^-$ ions.
- $NH4Cl (aq) \rightarrow NH4^+ (aq) + Cl^- (aq)$
The $NH_4^+$ ion of acidic salt undergoes hydrolysis.
In aqueous solution, the $NH4^+$ ion donates a proton to $H2O$ to form $NH3$ and $H3O^+$ ions.
- $NH4^+ (aq) + H2O (l) \rightleftharpoons NH3 (aq) + H3O^+ (aq)$
Thus, the solution is acidic (acidic salt).

Given that $Kb$ for $NH3$ is $1.8 \times 10^{-5}$ , calculate the pH of a 0.20 M $NH_4Cl$ solution.
$NH4Cl (aq) \rightarrow NH4^+ (aq) + Cl^- (aq)$
- Initial: 0.20 M 0 M 0 M
Hydrolysis of $NH_4^+$ :
- $NH4^+ (aq) + H2O (l) \rightleftharpoons NH3 (aq) + H3O^+ (aq)$
- Initial: 0.20 M 0 M 0 M
- Change: -x +x +x
- Final: (0.20 – x) M x M x M
$Ka = \frac{Kw}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$
$Ka = \frac{[H3O^+][NH3]}{[NH4^+]} = \frac{(x)(x)}{(0.20 - x)} \approx \frac{x^2}{0.20}$
$x = \sqrt{K_a \times 0.20} = \sqrt{5.56 \times 10^{-10} \times 0.20} = 1.05 \times 10^{-5}$
- Assume that x is small (0.2- x) 0.2
$[H_3O^+] = 1.05 \times 10^{-5} M$
$pH = -log[H_3O^+] = -log(1.05 \times 10^{-5}) = 4.98$

Basic salts are formed from the neutralisation reaction of weak acids (WA) react with strong bases (SB).
The pH of aqueous solutions of basic salts is greater than 7 (pH > 7).
The anion of basic salts will undergo hydrolysis.
Examples: $CH3COONa$ , $KNO2$
$CH3COOH (aq) + NaOH (aq) \rightarrow CH3COONa (aq) + H_2O (l)$
- Weak Acid + Strong Base -> Basic Salt
In aqueous solution, $CH3COONa$ dissociates completely to $Na^+$ and $CH3COO^-$ ions.
- $CH3COONa (aq) \rightarrow Na^+ (aq) + CH3COO^- (aq)$
The $CH_3COO^-$ ion of basic salt undergoes hydrolysis.
In aqueous solution, the $CH3COO^-$ ion accepts a proton from $H2O$ to form $CH_3COOH$ and $OH^-$ ions.
- $CH3COO^- (aq) + H2O (l) \rightleftharpoons CH_3COOH (aq) + OH^- (aq)$
Thus, the solution is basic (basic salt).

0. 05 mol sodium benzoate ( $C6H5COONa$ ) was dissolved in water in a 100 cm3 volumetric flask. What is the pH of the aqueous solution formed?
[ $Ka$ for $C6H_5COOH$ is $6.5 \times 10^{-5}$ ]
Concentration of $C6H5COONa = \frac{0.05 mol}{100 cm^3} \times \frac{1000 cm^3}{1 dm^3} = 0.5 mol/dm^3 = 0.5 M$
$C6H5COONa (aq) \rightarrow C6H5COO^- (aq) + Na^+ (aq)$
- Initial: 0.5 M 0 M 0 M
Hydrolysis of $C6H5COO^-$ :
- $C6H5COO^- (aq) + H2O (l) \rightleftharpoons C6H_5COOH (aq) + OH^- (aq)$
- Initial: 0.5 M 0 M 0 M
- Change: -x +x +x
- Final: (0.5 – x) M x M x M
$Kb = \frac{Kw}{K_a} = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}$
$Kb = \frac{[C6H5COOH][OH^-]}{[C6H_5COO^-]} = \frac{(x)(x)}{(0.5 - x)} \approx \frac{x^2}{0.5}$
$x = \sqrt{K_b \times 0.5} = \sqrt{1.54 \times 10^{-10} \times 0.5} = 8.77 \times 10^{-6}$
- Assume that x is small (0.5- x) 0.5
$[OH^-] = 8.77 \times 10^{-6} M$
$pOH = -log[OH^-] = -log(8.77 \times 10^{-6}) = 5.06$
$pH = 14 - pOH = 14 - 5.06 = 8.94$

The neutralization reactions of weak acids (WA) with weak bases (WB) may produce neutral, basic, or acidic salt.
- Neutral salt – ( $Ka = Kb$ ) – $Ka$ approximately equal to $Kb$ .
- Acidic salt – ( $Ka > Kb$ ) – Cation hydrolysis will be more extensive than anion hydrolysis.
- Basic salt – ( $Ka < Kb$ ) – Anion hydrolysis will be more extensive than cation hydrolysis.

$C6H5COONH4$ (ammonium benzoate) is an acidic salt because $Ka$ for $C6H5COOH$ ( $6.5 \times 10^{-5}$ ) is greater than $Kb$ for $NH3$ ( $1.8 \times 10^{-5}$ ).

Calculate the pH of 0.10 M $N2HCl$ solution. ( $K1$ for $N2H4 = 1.7 \times 10^{-7}$ )
Calculate the pH of 0.42 M $NH4NO3$ solution. ( $Kb$ for $NH3 = 1.8 \times 10^{-5}$ )
Calculate the pH of 0.25 M $CH3NH3NO3$ solution. ( $Kb$ for $CH3NH2 = 4.4 \times 10^{-4}$ )
Calculate the pH of a 0.15 M sodium acetate solution ( $CH3COONa$ ). ( $Ka$ for $CH_3COOH$ is $1.8 \times 10^{-5}$ )
Calculate the pH of a 0.24 M potassium formate solution ( $HCOOK$ ). ( $K_a$ for $HCOOH$ is $1.7 \times 10^{-4}$ )
Calculate the pH of a 0.75 M potassium hypochlorite solution ( $KClO$ ). ( $K_a$ for $HClO$ is $3.0 \times 10^{-8}$ )
Identify type of salt for the ammonium ethanoate solutions. (a: $Ka = 1.8 \times 10^{-5}$ , b: $Kb = 1.8 \times 10^{-5}$ )
Identify type of salt for the ammonium nitrite solutions. (a: $Ka = 4.2 \times 10^{-4}$ , b: $Kb = 1.8 \times 10^{-5}$ )
Predict the pH of the following solutions will be acidic, basic, or neutral:
- (a) $NH_4I$
- (b) $CaCl_2$
- (c) $KCN$
- (d) $NaClO_4$
- (e) $CH3NH3Br$
- (f) $HCOOK$