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))`

Options

A. `8.85xx10^(-3)`
B. `8.75xx10^(-2)`
C. `8.85xx10^(-4)`
D. `8.85xx10^(-2)`

Answer

A

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,a,0,0),("Mole after reaction", (0.01-a),0,a,a):}`
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-alpha)/(alpha)`
`:. a=8.85xx10^(-3)mol es of HCI`