Question 46 of 69intermediate🔧 ApplyNumerical3 marks

The ionization constant of acetic acid is 1.74×10−51.74 \times 10^{-5}. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Correct Answer

(i) Degree of dissociation (α\alpha) = 1.866×10−21.866 \times 10^{-2}

(ii) Concentration of acetate ion = 9.33×10−49.33 \times 10^{-4} M

(iii) pH = 3.03

Exercise: EXERCISES | Q: 6.46 | (Page 46)
For More Understanding

Explanation

The dissociation equilibrium of acetic acid is used to find the concentration of H+H^+ and Ac−Ac^- ions. Since the ionization constant KaK_a is very small, the change in concentration xx is negligible compared to the initial concentration, simplifying the calculation.

Solution Steps

  1. Step 1: Write the dissociation equilibrium For acetic acid, HAc(aq)⇌H+(aq)+Ac−(aq)\text{HAc(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{Ac}^-(\text{aq}) Initial concentration (M): 0.05000.05 \quad 0 \quad 0 Change (M): −x+x+x-x \quad +x \quad +x Equilibrium concentration (M): 0.05−xxx0.05-x \quad x \quad x

    Ka=[H+][Ac−][HAc]=x×x0.05−xK_a = \frac{[\text{H}^+][\text{Ac}^-]}{[\text{HAc}]} = \frac{x \times x}{0.05-x}

    Since KaK_a is small, x<<0.05x << 0.05, so 0.05−x≈0.050.05-x \approx 0.05.

  2. Step 2: Calculate the degree of dissociation

    1.74×10−5=x20.051.74 \times 10^{-5} = \frac{x^2}{0.05} x2=1.74×10−5×0.05=8.7×10−7x^2 = 1.74 \times 10^{-5} \times 0.05 = 8.7 \times 10^{-7}

    x=8.7×10−7=9.33×10−4x = \sqrt{8.7 \times 10^{-7}} = 9.33 \times 10^{-4} M Degree of dissociation, α=x0.05=9.33×10−40.05=1.866×10−2\alpha = \frac{x}{0.05} = \frac{9.33 \times 10^{-4}}{0.05} = 1.866 \times 10^{-2}

  3. Step 3: Calculate the concentration of acetate ion From the equilibrium, [Ac−]=x=9.33×10−4[\text{Ac}^-] = x = 9.33 \times 10^{-4} M

  4. Step 4: Calculate the pH of the solution [H+]=x=9.33×10−4[\text{H}^+] = x = 9.33 \times 10^{-4} M

    pH=−log⁡[H+]=−log⁡(9.33×10−4)=4−log⁡(9.33)=4−0.97=3.03\text{pH} = -\log[\text{H}^+] = -\log(9.33 \times 10^{-4}) = 4 - \log(9.33) = 4 - 0.97 = 3.03
PreviousNext