Question

What is the final pH of the solution that results when 10.0 mL of 0.1 M HCl(aq) is added to 90.0 mL of a buffer made of 1.00M HCN (Ka = 6.2 × 10-10) and 1.00M NaCN?.

Answer 1

mol of HCl added = 0.1M *10.0 mL = 1.0 mmol

CN- will react with H+ to form HCN

Before Reaction:

mol of CN- = 1.0 M *90.0 mL

mol of CN- = 90 mmol

mol of HCN = 1.0 M *90.0 mL

mol of HCN = 90 mmol

after reaction,

mol of CN- = mol present initially - mol added

mmol of CN- = (90 - 1.0) mmol

mol of CN- = 89 mmol

mol of HCN = mol present initially + mol added

mol of HCN = (90 + 1.0) mmol

mol of HCN = 91 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 {89/91}

= 9.198

pH is 9.20