Ionic Equilibria Notes

Salts

• 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

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

Relationship Between Dissociation Constants of Acids ($K<em>a$) and their Conjugate Base ($K</em>b$)

• The relationship between the dissociation constants of acids ($K<em>a$) and their conjugate base ($K</em>b$) can be derived as follows:
  • $CH_3COO^-$ hydrolysis in water:
    • Acid: $CH_3COOH$
    • Conjugate Base: $CH_3COO^-$
    • Base: $CH_3COO^-$
    • Conjugate Acid: $CH_3COOH$
• $K_w = [H^+][OH^-]$
• $K<em>a = \frac{[CH</em>3COO^-][H^+]}{[CH_3COOH]}$
• $K<em>b = \frac{[CH</em>3COOH][OH^-]}{[CH_3COO^-]}$
• $K<em>a \times K</em>b = \frac{[CH<em>3COO^-][H^+]}{[CH</em>3COOH]} \times \frac{[CH<em>3COOH][OH^-]}{[CH</em>3COO^-]} = [H^+][OH^-] = K_w$
  • Therefore, $K<em>a \times K</em>b = K_w$

Neutral Salts

• 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

• 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: $NH<em>4Cl$, $NH</em>4NO_3$
• $HCl (aq) + NH<em>3 (aq) \rightarrow NH</em>4Cl (aq)$
  • Strong Acid + Weak Base -> Acidic Salt
• In aqueous solution, $NH<em>4Cl$ dissociates completely to $NH</em>4^+$ and $Cl^-$ ions.
  • $NH<em>4Cl (aq) \rightarrow NH</em>4^+ (aq) + Cl^- (aq)$
• The $NH_4^+$ ion of acidic salt undergoes hydrolysis.
• In aqueous solution, the $NH<em>4^+$ ion donates a proton to $H</em>2O$ to form $NH<em>3$ and $H</em>3O^+$ ions.
  • $NH<em>4^+ (aq) + H</em>2O (l) \rightleftharpoons NH<em>3 (aq) + H</em>3O^+ (aq)$
• Thus, the solution is acidic (acidic salt).

Application: Calculating pH of an Acidic Salt Solution

• Given that $K<em>b$ for $NH</em>3$ is $1.8 \times 10^{-5}$, calculate the pH of a 0.20 M $NH_4Cl$ solution.
• $NH<em>4Cl (aq) \rightarrow NH</em>4^+ (aq) + Cl^- (aq)$
  • Initial: 0.20 M 0 M 0 M
• Hydrolysis of $NH_4^+$:
  • $NH<em>4^+ (aq) + H</em>2O (l) \rightleftharpoons NH<em>3 (aq) + H</em>3O^+ (aq)$
  • Initial: 0.20 M 0 M 0 M
  • Change: -x +x +x
  • Final: (0.20 – x) M x M x M
• $K<em>a = \frac{K</em>w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$
• $K<em>a = \frac{[H</em>3O^+][NH<em>3]}{[NH</em>4^+]} = \frac{(x)(x)}{(0.20 - x)} \approx \frac{x^2}{0.20}$
• $x = \sqrt{K_a \times 0.20} = \sqrt{5.56 \times 10^{-10} \times 0.20} = 1.05 \times 10^{-5}$
  • 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

• 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: $CH<em>3COONa$, $KNO</em>2$
• $CH<em>3COOH (aq) + NaOH (aq) \rightarrow CH</em>3COONa (aq) + H_2O (l)$
  • Weak Acid + Strong Base -> Basic Salt
• In aqueous solution, $CH<em>3COONa$ dissociates completely to $Na^+$ and $CH</em>3COO^-$ ions.
  • $CH<em>3COONa (aq) \rightarrow Na^+ (aq) + CH</em>3COO^- (aq)$
• The $CH_3COO^-$ ion of basic salt undergoes hydrolysis.
• In aqueous solution, the $CH<em>3COO^-$ ion accepts a proton from $H</em>2O$ to form $CH_3COOH$ and $OH^-$ ions.
  • $CH<em>3COO^- (aq) + H</em>2O (l) \rightleftharpoons CH_3COOH (aq) + OH^- (aq)$
• Thus, the solution is basic (basic salt).

Application: Calculating pH of a Basic Salt Solution

• 0. 05 mol sodium benzoate ($C<em>6H</em>5COONa$) was dissolved in water in a 100 cm3 volumetric flask. What is the pH of the aqueous solution formed?
• [$K<em>a$ for $C</em>6H_5COOH$ is $6.5 \times 10^{-5}$]
• Concentration of $C<em>6H</em>5COONa = \frac{0.05 mol}{100 cm^3} \times \frac{1000 cm^3}{1 dm^3} = 0.5 mol/dm^3 = 0.5 M$
• $C<em>6H</em>5COONa (aq) \rightarrow C<em>6H</em>5COO^- (aq) + Na^+ (aq)$
  • Initial: 0.5 M 0 M 0 M
• Hydrolysis of $C<em>6H</em>5COO^-$:
  • $C<em>6H</em>5COO^- (aq) + H<em>2O (l) \rightleftharpoons C</em>6H_5COOH (aq) + OH^- (aq)$
  • Initial: 0.5 M 0 M 0 M
  • Change: -x +x +x
  • Final: (0.5 – x) M x M x M
• $K<em>b = \frac{K</em>w}{K_a} = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}$
• $K<em>b = \frac{[C</em>6H<em>5COOH][OH^-]}{[C</em>6H_5COO^-]} = \frac{(x)(x)}{(0.5 - x)} \approx \frac{x^2}{0.5}$
• $x = \sqrt{K_b \times 0.5} = \sqrt{1.54 \times 10^{-10} \times 0.5} = 8.77 \times 10^{-6}$
  • Assume that x is small (0.5- x) 0.5
• $[OH^-] = 8.77 \times 10^{-6} M$
• $pOH = -log[OH^-] = -log(8.77 \times 10^{-6}) = 5.06$
• $pH = 14 - pOH = 14 - 5.06 = 8.94$

Salts of Weak Acids and Weak Bases

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

Example

• $C<em>6H</em>5COONH<em>4$ (ammonium benzoate) is an acidic salt because $K</em>a$ for $C<em>6H</em>5COOH$ ($6.5 \times 10^{-5}$) is greater than $K<em>b$ for $NH</em>3$ ($1.8 \times 10^{-5}$).

Exercises

• Calculate the pH of 0.10 M $N<em>2HCl$ solution. ($K</em>1$ for $N<em>2H</em>4 = 1.7 \times 10^{-7}$)
• Calculate the pH of 0.42 M $NH<em>4NO</em>3$ solution. ($K<em>b$ for $NH</em>3 = 1.8 \times 10^{-5}$)
• Calculate the pH of 0.25 M $CH<em>3NH</em>3NO<em>3$ solution. ($K</em>b$ for $CH<em>3NH</em>2 = 4.4 \times 10^{-4}$)
• Calculate the pH of a 0.15 M sodium acetate solution ($CH<em>3COONa$). ($K</em>a$ 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: $K<em>a = 1.8 \times 10^{-5}$, b: $K</em>b = 1.8 \times 10^{-5}$)
• Identify type of salt for the ammonium nitrite solutions. (a: $K<em>a = 4.2 \times 10^{-4}$, b: $K</em>b = 1.8 \times 10^{-5}$)
• 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) $CH<em>3NH</em>3Br$
  • (f) $HCOOK$