Question

4.Calculate the pH of a 0.500-L solution that contains 0.15 M formic acid, HCOOH (Ka =1.8 x 10^-4), and 0.20 M sodium formate, HCOONa. Then calculate the pH of the solution after the addition of 10.0 mL of 12.0 M NaOH.

Answer 1

1)

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

use:

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

= 3.745+ log {0.22/0.15}

= 3.911

Answer: 3.911

2)

mol of NaOH added = 12.0M *10.0 mL = 120.0 mmol

HCOOH will react with OH- to form HCOO-

Before Reaction:

mol of HCOO- = 0.2 M *500.0 mL

mol of HCOO- = 100 mmol

mol of HCOOH = 0.15 M *500.0 mL

mol of HCOOH = 75 mmol

100 mmol of HCOO- and NaOH would react

mol of NaOH remaining = 20 mmol

total volume = 500 mL + 10 mL

= 510 mL

[NaOH] = 20 mmol / 510 mL

= 0.03922 M

use:

pOH = -log [OH-]

= -log (3.922*10^-2)

= 1.4065

use:

PH = 14 - pOH

= 14 - 1.4065

= 12.5935

Answer: 12.59