Question

20. Calculate pH when 25.00 mL of 0.10 M HCOOH (Ka = 1.8*10^-4) is titrated with 5.00 mL of 0.25 M KOH?

Answer 1

first find the pKa value

pKa = logKa = -log1.8 x 10^-4 = 3.74

moles of HCOOH = 0.1 M x 0.025 L = 0.0025 mol

moles of KOH = 0.25 M x 0.005L = 0.00125 mol

let see the balanced equation

HCOOH + KOH ------> HCOOK + H2O

1 mol of HCOOH and one mole KOH react and form the 1 mol of HCOOK accordingly

0.00125 mol react with 0.00125 mol will give 0.00125 mol KOH will give 0.00125 mol of HCOOK

moles og HCOOH remaining = 0.0025 - 0.00125 = 0.00125 mol HCOOH remaining

total volume = 30 mL = 0.03L

concentration of HCOOK = 0.00125 mol / 0.03 L = 0.04167 M

moles of HCOOH = 0.00125 / 0.03 = 0.04167 mol / L

use hander son equation

pH = pKa + log[HCOOK / HCOOH]

pH = 3.74 + log[0.4167/0.04167]

pH = 3.74 + 0

pH = 3.74