First slide

Basic Buffer

Question

What amount of HCl will be required to prepare one litre of a buffer solution of pH 10.4 using 0.01 mole of NaCN ? Given $K_{ion} (HCN) = 4.1 \times 10^{- 10} .$

Moderate

A

$8.55 \times 10^{- 3} moles$
B

$8.65 \times 10^{- 3} moles$
C

$8.75 \times 10^{- 3} moles$
D

$9.9 \times 10^{- 4} moles$

Solution

The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have
[NaCN] = (0.01- x)
[HCN] = x
Substitution these values along with pH and $K_{a}$ in the expression.
$pH = - \log K_{a} + \log \frac{[Salt]}{[Acid]}$
We get 10.4 $= - \log [4 \times 10^{- 10}] + \log \frac{0.01 - x}{x}$
$or 10.4 = 9.4 + \log \frac{0.01 - x}{x}$
or $\log \frac{0.01 - x}{x} = 1$
or $\frac{0.01 - x}{x} = 10 \Rightarrow 11 x = 10^{- 2}$
or $x = 9.9 \times 10^{- 4} M$
So, number of moles $= 9.9 \times 10^{- 4}$

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