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 10ml. of 2M HCN?

1 Answer

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.