How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCN) of pH 8.5 using 0.01g formula weight of NaCN?

Explanation

Suppose, x mol of HCl is required to prepare the given buffer solution. Adding x mol of HCl to 1L of solution containing NaCN results in x mol of HCN by the reaction of HCl with NaCN
so, in the solution after addition of HCl
`[NaCN]=(0.1-x)mol L^(-1)` and [HCN] `=xmol*L^(-1)`
In case of a buffer solution consisting of NaCN and HCN,
`pH=pK_a+log([NaCN])/([HCN])`
Now, `pK_a=-log_(10)K_a=-log(4.1xx10^(-10))=9.38`
Given that pH of the buffer solution=8.5
`:.8.5=9.38+log(0.01-x)/(x)` or `x=8.85xx10^(-3)`
So, the amount of HCl necessary`=8.85xx10^(-3)mol`