#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#?

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#.