Question

3.Calculate the pH of a solution that contains 0.250 M formic acid, HCOOH (Ka=1.8 x 10^-4), and 0.100M sodium formate, HCOONa after the addition of 10.0 mL of 6.00M NaOH to the original buffered solution volume of 500.0 mL.

Answer 1

mol of NaOH added = 6.0M *10.0 mL = 60.0 mmol

HCOOH will react with OH- to form HCOO-

Before Reaction:

mol of HCOO- = 0.1 M *500.0 mL

mol of HCOO- = 50 mmol

mol of HCOOH = 0.25 M *500.0 mL

mol of HCOOH = 125 mmol

after reaction,

mol of HCOO- = mol present initially + mol added

mol of HCOO- = (50 + 60.0) mmol

mol of HCOO- = 110 mmol

mol of HCOOH = mol present initially - mol added

mol of HCOOH = (125 - 60.0) mmol

mol of HCOOH = 65 mmol

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

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

use:

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

= 3.745+ log {1.1*10^2/65}

= 3.973

Answer: 3.97