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Basic Buffer Action

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Question

Calculate the ratio of pH of a solution containing $1$ mole of ${C H}_{3} C O O N a + 1$ of $H C l$ per litre and of the other solution containing $1$ mole of ${C H}_{3} C O O N a + 1$ mole of ${C H}_{3} C O O H$ per litre.

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Solution

1st case:
${C H}_{3} C O O N a + H C l ⟶ {C H}_{3} C O O H + N a C l$
As we know for weak acid $p H = \frac{1}{2} (p K_{a} - l o g C)$
C=1 M
So $p H_{1} = - \frac{1}{2} log K_{a}$
2nd case:
$p H_{2} = - log K_{a} + log \frac{[S a l t]}{[A c i d]}$
[Salt]=[Acid]=1M
So $p H_{2} = - log K_{a}$
Ratio of $p H 1 : p H 2 = 1 : 2$

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