`K_(a)` for `HCN` is `5 xx 10^^(-10)` at `25^(@)C`. For maintaining a constant `pH` of `9.0`, the volume of `5M KCN` solution required to be added to `10mL` of `2M HCN` solution is

Options

A. `9.3 mL`
B. `7.95 mL`
C. `4mL`
D. `2mL`

Answer

D

Explanation

`pK_(a) =- log(5 xx 10^(-10))`
`=- 0.7 + 10 =9.3`
Acidic buffer will be formed.
`pH = pK_(a) + "log" ([KCN])/([HCN])`
Let `V mL` of `KCN` is added
Total volume `= (V +10)mL`
mmol of `KCN = 5 xx V`
`:. [KCN] = (5 xx V "mmol")/((V + 10)mL)`
mmol of `HCN = 2 xx 10 = 20`
`:. [HCN] = (20 "mmol")/((V +10)mL)`
Thus, from equation (i),
`9 = 9.3 + log[(5V//(V+10))/(20//(V+10))]`
`- 0.3 = log ((5V)(20))`
`log (V)/(4) = - 0.3`
`(V)/(4) = "Antilog" (-0.3 + 1-1)`
`= "Antilog" (bar(1).7) = 5 xx 10^(-1)`
`V = 20 xx 10^(-1) = 2 mL`