Question

A L.OOL buffer solution has the following concentrations: 1.50M formic acid and 1.25M sodium formate. The pK, for formic acid is 3.74. Determine the pH change of the buffer when 0.20moles of NaOH is added to the buffer. There is no change in the volume of the solution. (10 points)

Answer 1

Answer

∆pH = 0.12

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

pH = 3.74 + log ( 1.25M/1.50M)

pH = 3.74 - 0.079

pH = 3.66

Initial moles of HCOOH = 1.50 mol

Initial moles of HCOO^- = 1.25 mol

moles of NaOH added = 0.20moles

0.20moles of NaOH react with 0.20moles of HCOOH to produce 0.20moles of HCOO^-

HCOOH + OH^- --------> HCOO^-

After addition of NaOH

moles of HCOOH = 1.50mol - 0.20mol = 1.30mol

moles of HCOO^- = 1.25mol + 0.20mol = 1.45mol

[HCOOH] = 1.30M

[HCOO^-] = 1.45M

pH = pKa + log([A^-]/[HA])

pH = 3.74 + log( 1.45M/1.30M)

pH = 3.74 + 0.047

pH = 3.78

∆pH = 3.78 - 3.66 = 0.12