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  • What will be the value of pH of 0.01 mol dm-3 CH_{3}COOH(K_{a}1.74\times 10^{-5}) (i) 3.4 (ii) 3.6 (iii) 3.9 (iv) 3.0

What will be the value of pH of 0.01 mol dm-3 CH_{3}COOH(K_{a}=1.74\times 10^{-5})

(i) 3.4

 (ii) 3.6

 (iii) 3.9

 (iv) 3.0

 

Answers (1)

The answer is the option (i) 3.4

Explanation:  We know that the ionization of CH_{3}COOH occurs as follows: -

CH_{3}COOH+H_{2}O\leftrightharpoons H_{3}O^{+}+CH_{3}COO^{-}

Now, we know that the initial concentration: 0.01

We also know that the equilibrium concentration: 0.01 – x x x

Now, the ionization constant K_{a}=\frac{[H_{3}O^{+}][CH^{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.01-x}

As, x>>0.01 \; Therefore, 0.01 - \times \sim 0.01 .

X^{2}=1.74\times 10^{-5}\times 0.01

X = 4.2 \times 10^{-4}

Which can also be written as, pH = -log (4.2 \times 10^{-4} ) = 3.4

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infoexpert22

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