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# $K_a$ for HCN is $5 \times 10^{-10}$ at $25^\circ C$. How much volume of 5 M KCN solution is required to be added to 10 mL of 2 M HCN in order to maintain $pH$ = 9

(a) 2 mL

(b) 5 mL

(c) 3 mL

(d) 4 mL

Can you answer this question?

## 1 Answer

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Answer: 2 mL

$pH = pK_a + log \frac{[Salt]}{[Acid]}$
$\Rightarrow 9 = - log (5 \times 10^{-10}) + log \frac{[Salt]}{[Acid]}$
$\Rightarrow log \frac{[Salt]}{[Acid]} = 9 + log(5 \times 10^{-10}) = \bar1.6990$
$\Rightarrow \frac{[Salt]}{[Acid]} = antilog (1.6990) = 0.5$

Volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN in order to maintain pH = 9 is
$V = \frac{0.5 \times 2 M \times 10 mL \times 1000 }{5 M \times 1000} = 2 mL$

answered Nov 29, 2013
edited Nov 29, 2013

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