Question

Ka for hcn is 5*10^-10. For maintaining a constant ph of 9, the volume of 5 M kcn solution required to be added to 10ml of 2 M hcn solution is

Answer 1

$\text{use the formula for pH}$
$pH=pK_a+log\frac{[salt]}{[acid]}$

here, Ka = 5 × 10^-10
so, pKa = -log(5 × 10^-10) = -(log5 -log10)
pKa = -log5 + 10 and pH = 9
Let [ salt ]/[ acid ] = K
now,
pH = pKa + logK
9 = -log5 + 10 + logK
-1 + log5 = logK
$logK =\overline{1}.6990$
$K = \text{anilog}(\overline{1}.6990)$
K ≈ 0.5
hence, [ salt ]/[ acid ] = 0.5
means concentration of salt is 0.5 of concentration of acid .


now, volume of 5 M , KCN required to be added to 10 ml of 2M of HCN is V
[salt ] = 0.5 [ acid ]
MV = 0.5mv
V = 0.5mv/M
= 0.5 × 2M × 10ml/5 = 2ml

hence, volume of 5M , KCN is 2ml