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Question: What will be the value of pH of \(0.01moldm^{- 3}\) \[CH_{3}COOH(K_{a} = 1.74 \times 10^{- 5})?\]...

What will be the value of pH of 0.01moldm−30.01moldm^{- 3}

CH3COOH(Ka=1.74×10−5)?CH_{3}COOH(K_{a} = 1.74 \times 10^{- 5})?

A

3.43.4

B

3.63.6

C

3.93.9

D

3.03.0

Answer

3.43.4

Explanation

Solution

: CH3COOH+H2OCH_{3}COOH + H_{2}O

CH3COO−+H3O+CH_{3}COO^{-} + H_{3}O^{+}

Ka=[CH3COO−][H3O+][CH3COOH]K_{a} = \frac{\left\lbrack CH_{3}COO^{-} \right\rbrack\left\lbrack H_{3}O^{+} \right\rbrack}{\left\lbrack CH_{3}COOH \right\rbrack}

Ka=[H3O+]2[CH3COOH]K_{a} = \frac{\left\lbrack H_{3}O^{+} \right\rbrack^{2}}{\left\lbrack CH_{3}COOH \right\rbrack} (∵[CH3COO−]=[H3O+])\left( \because\left\lbrack CH_{3}COO^{-} \right\rbrack = \left\lbrack H_{3}O^{+} \right\rbrack \right)

[H3O+]2=Ka[CH3COOH]\left\lbrack H_{3}O^{+} \right\rbrack^{2} = K_{a}\left\lbrack CH_{3}COOH \right\rbrack

[H3O+]=Ka[CH3COOH]\left\lbrack H_{3}O^{+} \right\rbrack = \sqrt{K_{a}\left\lbrack CH_{3}COOH \right\rbrack}

Given:

Ka=1.74×10−5,[CH3COOH]=0.01moldm−3K_{a} = 1.74 \times 10^{- 5},\left\lbrack CH_{3}COOH \right\rbrack = 0.01moldm^{- 3}

[H3O+]=1.74×10−5×0.01=1.74×10−7\left\lbrack H_{3}O^{+} \right\rbrack = \sqrt{1.74 \times 10^{- 5} \times 0.01} = \sqrt{1.74 \times 10^{- 7}}

[H3O+]=4.17×10−4\left\lbrack H_{3}O^{+} \right\rbrack = 4.17 \times 10^{- 4}

pH=−log⁡[H3O+]=−log⁡(4.17×10−4)pH = - \log\left\lbrack H_{3}O^{+} \right\rbrack = - \log\left( 4.17 \times 10^{- 4} \right)

=3.379≈3.4= 3.379 \approx 3.4