Calculate the $P^{H}$ of the following mixtures given $P k_{a} = P k_{b} = 4.74$
$(1) 50 m l 0.1 M N a O H + 50 m l 0.1 M C H_{3} C O O H$
$(2) 50 m l 0.1 M N a O H + 50 m l 0.05 M C H_{3} C O O H$
$(3) 50 m l 0.05 M N a O H + 50 m l 0.1 M C H_{3} C O O H$

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8.72, 1.60, 4.74

7.62, 1.50, 3.54

8.32, 3.60, 6.74

9.52, 10.52, 6.51

answer is A.

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Detailed Solution

$(1)$ $\begin{array}{l} N a O H + C H_{3} C O O H \to C H_{3} C O O N a + H_{2} O \\ 50 m l, 0.1 M 50 m l, 0.1 M 0 \\ 5 m E q 5 m E q 0 \\ 0 0 5 m E q \\ [C H_{3} C O O N a] = [S a l t] = \frac{5}{100} = 0.05 M \\ f o r s a l t h y d r o l y s i s P^{H} = \frac{1}{2} (P_{k w} + P_{k a} + \log C) \\ f o r u n i v a l e n t s a l t = \frac{1}{2} (14 + 4.74 + \log (0.05)) \\ = \frac{1}{2} (14 + 4.74 - 2 + 0.7) = 8.72 \end{array}$

$(2) [N a O H] = [O H^{-}] = \frac{2.5}{100} = 2.5 \times 10^{- 2} M$

For base, $\begin{array}{l} P^{O H} = - \log [O H^{-}] = - \log 2.5 \times 10^{- 2} \\ = - \log 2.5 + 2 \\ = - 4 + 2 = 1.60 \end{array}$

$(3)$ For buffer solution $\begin{array}{l} P^{H} = P_{k a} + \log \frac{[S]}{[A]} \\ = 4.74 + \log \frac{[2.5]}{[2.5]} \\ = 4.74 \end{array}$

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