Ncert Exercises & Solutions

The value of the pH of \(0.01 \) \(\text{mol dm}^{-3} \)\(\text{CH}_3\text{COOH}\) \(\left(K_a=1.74 \times 10^{-5}\right)\) 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