Question

A certain volume of 1M NaOH is needed to titrate 500mL of a 0.2M solution of a weak acid(pKa=4.0) to the equivalence point. You then titrate a 0.4M solution of this same weak acid but to pH=4.0. If you used the same amount of 1.0 M NaOH for the second titration as for the first, the volume of 0.4M solution titrated is

A)100mL

B) 250mL

C)500mL

D)1000mL

E)200mL

Explain and show all work.

Answer 1

Case 1:

Molarity of acid solution = 0.2 M

Volume of solution titrated = 500 mL = 0.5 L

pKa of acid = 4.0

Ka = 10^-4

Ka = [H+][A-] / [HA]

[H+] =

=

[H+] = 0.00447 M

Moles of NaOH required = 0.00447 * 0.5

= 0.002235

Volume of NaOH needed = 0.002235 / 1 = 0.002235 L

= 2.235 mL

Case 2:

Molarity of acid solution = 0.4 M

pH of solution = 4

Since same volume of NaOH is used:

Moles of NaOH used = 0.002235

Moles of H+ neutralized = 0.002235

pH of solution after titration = 4

[H+] left in solution = 10^-4 M

[H+] from acid initially =

= 0.0063 M

Let the volume of acid solution be X L

Total moles of H+ given by acid solution = 0.0063 * X

Moles of H+ left in the solution after nutralization = 0.0063X - 0.002235

Total volume of solution = (X + 0.002235) L

[H+] left in the solution = (0.0063X - 0.002235) / (X + 0.002235)

10^-4 = (0.0063X - 0.002235) / (X + 0.002235)

X = 500 mL