Question

A 50.0mL solution contains 0.120M HCN (Ka=6.2x10^-10) and 0.240M NaCN

Calculate the final pH of the solution after adding 20.0 mL of 0.0800 M HCl

Answer 1

mol of HCl added = 0.0800 M *20.0 mL = 1.60 mmol

CN- will react with H+ to form HCN

Before Reaction:

mol of CN- = 0.240 M *50.0 mL

mol of CN- = 12.0 mmol

mol of HCN = 0.120 M *50.0 mL

mol of HCN = 6.00 mmol

after reaction,

mol of CN- = mol present initially - mol added

mmol of CN- = (12.0 - 1.60) mmol

mol of CN- = 10.4 mmol

mol of HCN = mol present initially + mol added

mol of HCN = (6.00 + 1.60) mmol

mol of HCN = 7.60 mmol

Ka = 6.2*10^-10

pKa = - log (Ka)

= - log(6.2*10^-10)

= 9.2076

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

use:

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

= 9.2076+ log {10.4/7.6}

= 9.34

pH is 9.34