`K_(a)` for HCN is `5.0xx10^(-10)` at `25^(@)C`. For maintaining a constant pH of 9. Calculate the volume of `5.0M KCN` solution required to be added to 10 mL of `2.0M HCN` solution.

Options

A. 2 mL
B. 5 mL
C. 3mL
D. 4mL

Answer

A

Explanation

`pH=pK_(a)+log.(["Salt"])/(["Acid"])`
`9=-log (5xx10^(-10))+log.(["Salt"])/(["Acid"])`
or `log. (["Salt"])/(["Acid"])=9+log (5xx10^(-10))=1.6990`
or `(["Salt"])/(["Acid"])=`antilog`(1.6990)=0.5`
Ph `=-log K_(a)+log.(["Salt"])/(["Acid"])`