The equilibrium constant (K_C) for the reaction of a weak acid HA with strong base NaOH is 10 at $25^{0} C$ . Which of the following statement(s) is/are correct ?

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The ionization constant (Ka) at $25^{0} C$ is $10^{- 5}$ for HA

If K_b of a weak base BOH is $10^{- 4}$ at $25^{0} C$ .equilibrium constant for neutralization of HA with BOH at $25^{0} C$ will be $10^{5}$

pH of a 0.10M aqueous solution of NaA at $25^{0} C$ will be 9

pH of a 0.01M aqueous solution of HA at $25^{0} C$ will be 3.5

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} H A + N a O H ⇌ N a A + H_{2} O; K = 10^{9} \\ (w e a k) \end{array}$

$K = \frac{1}{K h} \Rightarrow K h = 10^{- 9}$

A) $K h = \frac{K w}{K a} (o r) K a = \frac{K w}{K h} = \frac{10^{- 14}}{10^{- 9}} = 10^{- 5}$

(B) pH of 0.01M $H A = \frac{1}{2} p K a - \frac{1}{2} \log C = 3.5$

(D) $K = \frac{1}{K h} = \frac{K a \times K b}{K w} = \frac{10^{- 5} \times 10^{- 4}}{10^{- 14}} = 10^{5}$

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