Question

a) Calculate the pH for a 1.63 M solution of HCN which also contains 1.75 M NaCN. Ka = 4.9 x 10-10 for HCN b) Calculate the pH after 65.0 mL of 1.47 M HCl is added to 850.0 mL of the solution in part a) above.

Answer 1

a)

Ka = 4.9*10^-10

pKa = - log (Ka)

= - log(4.9*10^-10)

= 9.31

use:

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

= 9.31+ log {1.75/1.63}

= 9.341

Answer: 9.34

b)

mol of HCl added = 1.47M *65.0 mL = 95.55 mmol

CN- will react with H+ to form HCN

Before Reaction:

mol of CN- = 1.75 M *850.0 mL

mol of CN- = 1487.5 mmol

mol of HCN = 1.63 M *850.0 mL

mol of HCN = 1385.5 mmol

after reaction,

mol of CN- = mol present initially - mol added

mmol of CN- = (1487.5 - 95.55) mmol

mol of CN- = 1391.95 mmol

mol of HCN = mol present initially + mol added

mol of HCN = (1385.5 + 95.55) mmol

mol of HCN = 1481.05 mmol

Ka = 4.9*10^-10

pKa = - log (Ka)

= - log(4.9*10^-10)

= 9.31

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

use:

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

= 9.31+ log {1.392*10^3/1.481*10^3}

= 9.283

Answer: 9.28