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Catalytic Hydrogenation

500 mL of 0...

Question

$500$ mL of $0.01 M$ ${C H}_{3} C O O H + 500$ mL of $0.02 M$ ${C H}_{3} C O O N a$ gives a $p H$ equal to $5.3010$ . The $p K_{b}$ of ${C H}_{3} C O O^{-}$ is:

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Solution

$p H = p K_{a} + l o g \frac{C H_{3} C O O N a}{C H_{3} C O O H}$
$5.3010 = p K_{a} + l o g \frac{\frac{0.02}{0.5}}{\frac{0.01}{0.5}}$
$p K a = 5$
$p K_{b} = p K_{w} - p K_{a} = 14 - 5 = 9$

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