Question

4). A 25.0-mL sample of 0.265 M propanoic acid (Ka = 1.3x10-5) is titrated with 0.2500 M NaOH. What is the pH after 15mL of base was added?

Answer 1

Given:

M(HA) = 0.265 M

V(HA) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 15 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.265 M * 25 mL = 6.625 mmol

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

mol(NaOH) = 0.25 M * 15 mL = 3.75 mmol

We have:

mol(HA) = 6.625 mmol

mol(NaOH) = 3.75 mmol

3.75 mmol of both will react

excess HA remaining = 2.875 mmol

Volume of Solution = 25 + 15 = 40 mL

[HA] = 2.875 mmol/40 mL = 0.0719M

[A-] = 3.75/40 = 0.0938M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

use:

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

= 4.886+ log {9.375*10^-2/7.187*10^-2}

= 5.001

Answer: 5.00