Question

A titration is carried out using 20 mL of 0.500 M Formic Acid (1.8 x 10-4) with 0.500 M NaOH. Calculate the pH after 10.0 mL, 20.0 mL, and 30.0 mL of NaOH have been added.

Answer 1

1)when 10.0 mL of NaOH is added

Given:

M(HCOOH) = 0.5 M

V(HCOOH) = 20 mL

M(NaOH) = 0.05 M

V(NaOH) = 10 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.5 M * 20 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.05 M * 10 mL = 0.5 mmol

We have:

mol(HCOOH) = 10 mmol

mol(NaOH) = 0.5 mmol

0.5 mmol of both will react

excess HCOOH remaining = 9.5 mmol

Volume of Solution = 20 + 10 = 30 mL

[HCOOH] = 9.5 mmol/30 mL = 0.3167M

[HCOO-] = 0.5/30 = 0.0167M

They form acidic buffer

acid is HCOOH

conjugate base is HCOO-

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 {1.667*10^-2/0.3167}

= 2.466

2)when 20.0 mL of NaOH is added

Given:

M(HCOOH) = 0.5 M

V(HCOOH) = 20 mL

M(NaOH) = 0.05 M

V(NaOH) = 20 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.5 M * 20 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.05 M * 20 mL = 1 mmol

We have:

mol(HCOOH) = 10 mmol

mol(NaOH) = 1 mmol

1 mmol of both will react

excess HCOOH remaining = 9 mmol

Volume of Solution = 20 + 20 = 40 mL

[HCOOH] = 9 mmol/40 mL = 0.225M

[HCOO-] = 1/40 = 0.025M

They form acidic buffer

acid is HCOOH

conjugate base is HCOO-

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 {2.5*10^-2/0.225}

= 2.79

3)when 30.0 mL of NaOH is added

Given:

M(HCOOH) = 0.5 M

V(HCOOH) = 20 mL

M(NaOH) = 0.05 M

V(NaOH) = 30 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.5 M * 20 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.05 M * 30 mL = 1.5 mmol

We have:

mol(HCOOH) = 10 mmol

mol(NaOH) = 1.5 mmol

1.5 mmol of both will react

excess HCOOH remaining = 8.5 mmol

Volume of Solution = 20 + 30 = 50 mL

[HCOOH] = 8.5 mmol/50 mL = 0.17M

[HCOO-] = 1.5/50 = 0.03M

They form acidic buffer

acid is HCOOH

conjugate base is HCOO-

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 {3*10^-2/0.17}

= 2.991