Question

An analytical chemist is titrating 138.2 mL of a 1.100 M solution of butanoic acid (HC3H7CO2) with a 0.6000 M solution of KOH. The pKa of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 294.3 mL of the KOH solution to it. Please round to 2 decimal places.

Answer 1

use:

pKa = -log Ka

4.82 = -log Ka

Ka = 1.514*10^-5

Given:

M(HC3H7CO2) = 1.1 M

V(HC3H7CO2) = 138.2 mL

M(KOH) = 0.6 M

V(KOH) = 294.3 mL

mol(HC3H7CO2) = M(HC3H7CO2) * V(HC3H7CO2)

mol(HC3H7CO2) = 1.1 M * 138.2 mL = 152.02 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.6 M * 294.3 mL = 176.58 mmol

We have:

mol(HC3H7CO2) = 152.02 mmol

mol(KOH) = 176.58 mmol

152.02 mmol of both will react

excess KOH remaining = 24.56 mmol

Volume of Solution = 138.2 + 294.3 = 432.5 mL

[OH-] = 24.56 mmol/432.5 mL = 0.0568 M

use:

pOH = -log [OH-]

= -log (5.679*10^-2)

= 1.2458

use:

PH = 14 - pOH

= 14 - 1.2458

= 12.7542

Answer: 12.75