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)=4.1xx10^(-10)`.

Options

A. `8.55xx10^(-3)`moles
B. `8.65xx10^(-3)`moles
C. `8.75xx10^(-3)`moles
D. `9.9xx10^(-4)`moles

Answer

D

Explanation

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`
Substituting these values along with `pH` and `K_(a)` in the expression
`pH= -log K_(a)+log``(["Salt"])/([Acid])`
We get `10.4= -log[4xx10^(-10)]+log``(0.01-x)/(x)`
or `10.4=9.4+log'(0.01-x)/(x)`
or `log(0.01-x)/(x)=1`
or `(0.01-x)/(x)=10implies11x=10^(2)`
or `x=9.9xx10^(-4) M`
So, number of moles `=1xx9.9xx10^(-4)`
`=9.9xx10^(-4) mol`