Question

`K_a` for `HCN` is `5xx10^(-10)` at a certain temperature. For maintaining a constant `pH` of `9`, what is the volume of `5M KCN` solution required to be added to `10`ml. of `2M HCN`?

Answer 1

2mL.

Explanation:

KCN and HCN make a tampon solution with pH that you can calculate with the formule `pH = pKa + log ((Cs)/(Ca))` where `pKa=-log_10 Ka= -log5-log10^(-10)= 9.3` .
If pH must be 0 you have `log ((Cs)/(Ca))=pH - pKa = 9-9.3 =- 0.3`
therefore `(Cs)/(Ca)=10^(-0.3) = 0.5`.

Since `C_s= n°` mol of salt divided the volume and `C_a= n°` mol of acid divided the volume, and as the volume is the same you can write `(n_s/n_a)= 0.5` and you need as much salt as half of the acid.

mol of acid are :
`mol= M xx V= 2mol xx 0.01 L= 0.02 mol`
so you need 0.01 mol of salt whose volume is : `V=(0.01 mol)/((5 (mol)/L))= 0.002 L = 2 mL`.