The pH of 0-3 M acetic acid CH 3 COOH with K a -1-8 - 10-5 is -Take log 102-32-0-36A- 6-56B- 4-86C- 7-63D- 2-64

The correct option is D 2.64K_a=1.8×10^ 5 the dissociation of acetic acid is given by CH_3COOH ⇌ H^++CH_3COO^ Initial: 0.3 0 ...

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Standard XII

Chemistry

pH the Power of H

The pH of 0.3...

Question

The pH of $0.3 M$ acetic acid $(C H_{3} C O O H)$ with $K_{a} = 1.8 \times 10^{- 5}$ is :
Take $l o g_{10} (2.32) = 0.36$

6.56

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2.64

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4.86

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7.63

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Solution

The correct option is B 2.64
$K_{a} = 1.8 \times 10^{- 5}$
the dissociation of acetic acid is given by
$C H_{3} C O O H ⇌ H^{+} + C H_{3} C O O^{-} I n i t i a l : 0.3 00 A t e q u i l i b r i u m : 0.3 - x x x H e r e x i s t h e d i s s o c i a t i o n o f C H_{3} C O O H a t e q u i l i b r i u m$

$\Rightarrow K_{a} = 1.8 \times 10^{- 5} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = \frac{x \times x}{0.3 - x} S i n c e C H_{3} C O O H i s a w e a k a c i d x << 0.3$
$∴ 0.3 - x \approx 0.3 K_{a} = 1.8 \times 10^{- 5} = \frac{x^{2}}{0.3} \Rightarrow x = 2.32 \times 10^{- 3} x = [H^{+}] = 2.32 \times 10^{- 3} p H = - l o g_{10} [H^{+}] p H = - l o g_{10} (2.3 \times 10^{- 3}) p H = 2.638$

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pH the Power of H

Standard XII Chemistry

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The pH of 0.3 M acetic acid (CH3COOH) with Ka=1.8×10−5 is : Take log10(2.32)=0.36

The pH of $0.3 M$ acetic acid $(C H_{3} C O O H)$ with $K_{a} = 1.8 \times 10^{- 5}$ is :
Take $l o g_{10} (2.32) = 0.36$