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Common Ion Effect

When 100 mL...

Question

When $100 m L$ of $0.4 M C H_{3} C O O H$ are mixed with $100 m L$ of $0.2 M N a O H$ , the $[H_{3} O^{+}]$ in the solution is approximately
$[K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}]$

1.8 \times 10^{- 6}

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1.8 \times 10^{- 5}

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9 \times 10^{- 6}

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9 \times 10^{- 5}

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Solution

The correct option is B $1.8 \times 10^{- 5}$
$C H_{3} C O O H + N a O H ⟶ C H_{3} C O O N a + H_{2} O$
$100 m L, 0.4 M 100 m L, 0.2 M$
$4020$
After reaction,
$[C H_{3} C O O H] = \frac{20}{200}, [C H_{3} C O O N a] = \frac{20}{200}$

$p H = p K_{a} + log \frac{[s a l t]}{[a c i d]}$

$p H = p K_{a}$

$[H^{+}] = K_{a} = 1.8 \times 10^{- 5}$

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100 mL

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[K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}]

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H_{2} O . [K_{a (C H_{3} C O O H)} = 1.8 \times 10^{- 5}; K_{b (N H_{4} O H)} = 1.8 \times 10^{- 5}]

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