Question

Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka=1.3x10-5) after adding 20.98 mL of 0.1000M NaOH.

Answer 1

Given:

M(HPr) = 0.1 M

V(HPr) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 20.98 mL

mol(HPr) = M(HPr) * V(HPr)

mol(HPr) = 0.1 M * 40 mL = 4 mmol

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

mol(NaOH) = 0.1 M * 20.98 mL = 2.098 mmol

We have:

mol(HPr) = 4 mmol

mol(NaOH) = 2.098 mmol

2.098 mmol of both will react

excess HPr remaining = 1.902 mmol

Volume of Solution = 40 + 20.98 = 60.98 mL

[HPr] = 1.902 mmol/60.98 mL = 0.0312M

[Pr-] = 2.098/60.98 = 0.0344M

They form acidic buffer

acid is HPr

conjugate base is Pr-

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.8861

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.8861+ log {0.0344/0.0312}

= 4.93

Answer: 4.93