Question

What is the pH of the solution of 90 mL of 1.1 HCl is added to a solution of 250 mL of 1.0 M HCN and 1.0 NaCN? (HCN K_a = 6.2 x10^-10)

Answer 1

mol of HCl added = 1.1M *90.0 mL = 99.0 mmol

CN- will react with H+ to form HCN

Before Reaction:
mol of CN- = 1.0 M *250.0 mL
mol of CN- = 250 mmol

mol of HCN = 1.0 M *250.0 mL
mol of HCN = 250 mmol

after reaction,
mol of CN- = mol present initially - mol added
mmol of CN- = (250 - 99.0) mmol
mol of CN- = 151 mmol

mol of HCN = mol present initially + mol added
mol of HCN = (250 + 99.0) mmol
mol of HCN = 349 mmol


Ka = 6.2*10^-10

pKa = - log (Ka)
= - log(6.2*10^-10)
= 9.208

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.208+ log {1.51*10^2/3.49*10^2}
= 8.844

pH is 8.84