Question 49 of 69intermediate🔧 ApplyNumerical3 marks

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKapK_a of bromoacetic acid.

Correct Answer

(pH) 1.88

(pKa) 2.70

Exercise: EXERCISES | Q: 6.50 | (Page 46)
For More Understanding

Explanation

The degree of ionization (α\alpha) relates the initial concentration (cc) to the equilibrium concentration of H+H^+ ions via the formula [H+]=cα[H^+] = c\alpha. Using this, we find the pH. For a weak acid, the ionization constant KaK_a is given by Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}. Once KaK_a is calculated, pKapK_a is found using pKa=−log⁡(Ka)pK_a = -\log(K_a).

Solution Steps

  1. Step 1: Calculate the concentration of H+H^+ ions Given, concentration of bromoacetic acid, c=0.1c = 0.1 M Degree of ionization, α=0.132\alpha = 0.132 [H+]=c×α=0.1×0.132=0.0132[H^+] = c \times \alpha = 0.1 \times 0.132 = 0.0132 M

  2. Step 2: Calculate the pH of the solution

    pH=−log⁡[H+]pH = -\log [H^+] pH=−log⁡(0.0132)=−log⁡(1.32×10−2)pH = -\log(0.0132) = -\log(1.32 \times 10^{-2}) pH=2−log⁡(1.32)=2−0.12=1.88pH = 2 - \log(1.32) = 2 - 0.12 = 1.88

  3. Step 3: Calculate the ionization constant KaK_a For a weak acid, Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}

    Ka=0.1×(0.132)21−0.132=0.1×0.0174240.868K_a = \frac{0.1 \times (0.132)^2}{1 - 0.132} = \frac{0.1 \times 0.017424}{0.868} Ka=0.00174240.868=2.01×10−3K_a = \frac{0.0017424}{0.868} = 2.01 \times 10^{-3}

  4. Step 4: Calculate the pKapK_a of bromoacetic acid

    pKa=−log⁡(Ka)pK_a = -\log(K_a) pKa=−log⁡(2.01×10−3)=3−log⁡(2.01)pK_a = -\log(2.01 \times 10^{-3}) = 3 - \log(2.01) pKa=3−0.30=2.70pK_a = 3 - 0.30 = 2.70
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