0.01 gram-mole of NaCN is dissolved in 1L of an aqueous solution. How many gram-moles of HCl should be added to this solution in order to make it a buffer solution of pH= `8.5[K_a(HCN)=4.1xx10^(-10)]`

Explanation

NaCN reacts with HCl to form HCN.
`NaCN(aq)+HCl(aq)toHCN(aq)+NaCl(aq)`
If no. of moles of HCJ is less than that of NaCN, the solution of (NaCN + HCN) will be a buffer solution.
pH of this buffer solution `=pk_a+log([NaCN])/([HCN])`
`pK_a=-log_(10)(4.1xx10^(-10))=9.38`
Let, no. of gram-moles of HCI added to the solution = a gram-mol HCI reacts to give a gram-mol HCN. So, no. of gram-moles of NaCN left in the solution
=(0.01-a)
`:'` Volume of the solution=1L [HCN]=a `mol*L^(-1)`
and [NaCN]=(0.1-a)`mol*L^(-1)`
As given in the question, pH=8.5
`:.8.5=9.38+log((0.01-a)/a)` or `log((0.01-a)/a)=-0.88`
or `((0.01-a)/(a))=0.132` or `0.01-a=0.132a`
or, `a=(0.01)/(1.132)=8.83xx10^(-3)`
Therefore, `8.83xx10^(-3)` gram-mol of HCl should be
added in order to make the solution a buffer solution of pH= 8.5.