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One liter of $0.1 M$ of a weak acid $H {A (KK}_{a} = 2.0 \times 10^{-} M)$ is to be converted into a buffer of $p H = 4.4$ .The mass of $NaA (molar mass of NaA = 100 g {mol}^{- 1})$ to be added to this solution is $Given : \log 2 = 0.30)$ $( Given : \log 2 = 0.30)$

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$2.5 g$

$3.0 g$

$3.5 g$

$5.0 g$

answer is D.

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Detailed Solution

$p K_{a}^{\circ} = - \log (2.0 \times 10^{- 5} M) = 4.70$

Using the expression $pH = p K_{a}^{\circ} + \log {[salt] / [acid]}$

$\log (\frac{[acid]}{[salt]}) = 0.3$ This gives $\frac{[acid]}{[salt]} = 2$ $[salt] = [acid] / 2 = 0.1 M / 2 = 0.05 M$

Mass of $NaA$ required $= (0.05 mol) (100 g {mol}^{- 1}) = 5.0 g$

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