Question

500.0 mL of 0.140 M NaOH is added to 615 mL of 0.200 M weak acid (Ka = 4.06 × 10-5). What is the pH of the resulting buffer?

Answer 1

Given:

M(HA) = 0.2 M

V(HA) = 615 mL

M(NaOH) = 0.14 M

V(NaOH) = 500 mL

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

mol(HA) = 0.2 M * 615 mL = 123 mmol

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

mol(NaOH) = 0.14 M * 500 mL = 70 mmol

We have:

mol(HA) = 123 mmol

mol(NaOH) = 70 mmol

70 mmol of both will react

excess HA remaining = 53 mmol

Volume of Solution = 615 + 500 = 1115 mL

[HA] = 53 mmol/1115 mL = 0.0475M

[A-] = 70/1115 = 0.0628M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 4.06*10^-5

pKa = - log (Ka)

= - log(4.06*10^-5)

= 4.391

use:

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

= 4.391+ log {6.278*10^-2/4.753*10^-2}

= 4.512

Answer: 4.51