Question

A 1.0 L of a buffer solution is created which is 0.363 M in hydrocyanic acid, HCN, and 0.303 M sodium cyanate, NaCN. K_a for HCN = 4.0 x 10^-10.

What is the pH after 0.089 mol of HCl is added to the buffer solution?

Answer 1

mol of HCl added = 0.089 mol

CCN- will react with H+ to form HCN

Before Reaction:

mol of CCN- = 0.303 M *1.0 L

mol of CCN- = 0.303 mol

mol of HCN = 0.363 M *1.0 L

mol of HCN = 0.363 mol

after reaction,

mol of CCN- = mol present initially - mol added

mol of CCN- = (0.303 - 0.089) mol

mol of CCN- = 0.214 mol

mol of HCN = mol present initially + mol added

mol of HCN = (0.363 + 0.089) mol

mol of HCN = 0.452 mol

Ka = 4*10^-10

pKa = - log (Ka)

= - log(4*10^-10)

= 9.398

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.398+ log {0.214/0.452}

= 9.073

Answer: 9.07