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pH the Power of H

If pKb for ...

Question

If $p K_{b}$ for $C N^{-}$ at $25^{o}$ C is $4.7$ . The $p H$ of $0.5 M$ aqueous $N a C N$ solution is?

12

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10

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11.5

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11

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Solution

The correct option is C $11.5$
$C N^{-}$ will come from complete ionization of NaCN, Thus
Reaction can be represented as:

$C N^{-} + H_{2} O \to H C N + O H^{-}$
initial concentrations:
$0.5 00$
at equillibrium:
$0.5 - x x x$

given that $p K_{b} = 4.7$
as $p K_{b} = - l o g K_{b}$

$4.7 = - l o g K_{b}$
or $K_{b} = 2 \times 10^{- 5}$

since $K_{b} = \frac{[H C N] [O H^{-}]}{[C N^{-}]}$

$2 \times 10^{- 5} = \frac{x^{2}}{0.5 - x}$

x is very small thus $0.5 - x \approx 0.5$

$x = [O H^{-}] = 3.16 \times 10^{- 3}$

pOH=-log([ $O H^{-}$ ])
pOH=-log ( $3.16 \times 10^{- 3}$ )
pOH=2.5
using $p O H + p H = 14$
we get $p H = 14 - 2.5 = 11.5$

thus pH of the solution is 11.5.

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