Question

Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka=1.3x10-5) after adding 16.33 mL of 0.1000M NaOH. ** All volumes should have a minimum of 2 decimal places.

Answer 1

Given:
M(HPr) = 0.1 M
V(HPr) = 40 mL
M(NaOH) = 0.1 M
V(NaOH) = 16.33 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 * 16.33 mL = 1.633 mmol


We have:
mol(HPr) = 4 mmol
mol(NaOH) = 1.633 mmol

1.633 mmol of both will react

excess HPr remaining = 2.367 mmol
Volume of Solution = 40 + 16.33 = 56.33 mL
[HPr] = 2.367 mmol/56.33 mL = 0.042M

[Pr-] = 1.633/56.33 = 0.029M

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 {0.029/0.042}
= 4.72


Answer: 4.72