Ncert Exercises & Solutions

The value of pH of $0.01 mol {dm}^{- 3} {CH}_{3} COOH$ $(K_{a} = 1.74 \times 10^{- 5})$ is-

1. 3.4

2. 3.6

3. 3.9

4. 3.0

Hint:

[H^{+}] = \sqrt{K_{a} . C}

Step 1:

Calculate the concentration of H⁺ ion as follows:

Given that,

(K_{a} = 1.74 \times 10^{- 5})

Concentration of

{CH}_{3} COOH = 0.01 mol {dm}^{- 3}

[H^{+}] = \sqrt{K_{a} . C} = \sqrt{1.74 \times 10^{- 5} \times 0.01} = 4.17 \times 10^{- 4} Step 2 : pH = - \log [H^{+}] = - \log (4.17 \times 10^{- 4}) = 3.4