`K_a` for HCN is `5xx10^(-10)` at `25^(@)C`. For maintaining a constant pH=9, the volume of 5M KCN solution required to be added to 10mL of 2M HCN solution is-

Options

A. 4mL
B. 2.5 mL
C. 2mL
D. 6.4 mL

Explanation

`pH=pK_a+log(["salt])/(["acid"])`
`pK_a+log([KCN])/([HCN])`
Let the volume of KCN solution required be V mL
`:.[KCN]=(5xxV)/(V+10)` and `[HCN]=(10xx2)/(V+10)`
Now from equation (i)
pH`=-log(5xx10^(10))+log[(5xxV)/(V+10)//(10xx2)/(V+10)]`
`9=-log(5xx10^(-10))+logV/4`
On solving `V=1.99~~2mL`