How many mole of HCI will be required to prepare one litre of buffer solution (containing `NaCN+ HCI`) of `pH 8.5` using `0.01`g formula weight of NaCN ? `K_(HCN)=4.1xx10^(-10))`

Explanation

`NaCN+HCI` is not a buffer but if HCI is in less amount then it gives a buffer as it produces HCN.
`{:(,NaCN+,HCIrarr,NaCI+,HCN),("Mole added", 0.01,alpha,0,0),("Mole after reaction", (0.01-alpha),0,alpha,alpha):}`
This is buffer of HCN + NaCN.
Let a mole of HCI be used for this purpose.
`:. pH= -log K_(a)+log((0.01-a)/(a))`
`8.5= -log 4.1xx10^(-10)+log ((0.01-a)/(a))`
`:. a = 8.85xx10^(-3)` mole of HCI`