Calculate the $p O H$ and $p H$ of a 0.1 M $C H_{3} C O O H$ solution ( $K_{a}$ = $1.8 \times 10^{- 5}$ ). There is $7.5 \times 10^{- 3} %$ yield of $O H^{-}$ ions in the given reaction $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$

6.12, 7.88

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4.12, 9.88

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5.13, 8.87

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None of the options are correct

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Solution

The correct option is C
5.13, 8.87

$p H$ = $7 + \frac{1}{2} p K_{a}$ + $\frac{1}{2} l o g C$

= $7 + \frac{1}{2} \times 4.74$ + $\frac{1}{2} l o g (0.1)$

$p H$ = $8.87$ and $p O H$ = $14$ – $8.87$ = $5.13$

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Similar questions

C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}; K_{1} = 1.8 \times 10^{- 5}

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{2} = 1 \times 10^{14}

C H_{3} C O O H + O H^{-} ⇌ C H_{3} C O O^{-} + H_{2} O

x \times 10^{9}

K_{a}

2.0 \times 10^{- 5}

0.2 M

C H_{3} C O O^{-}

0.1 M

C H_{3} C O O H

K_{a} = 1.8 \times 10^{- 5}

C H_{3} C O O H + O H^{-} ⇋ C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} ⇋ H_{2} O + q_{2}

C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}

C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} \to H_{2} O + q_{2}

C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}

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