Question

500.0 mL of 0.140 M NaOH is added to 595 mL of 0.200 M weak acid (Ka = 2.29 × 10-5). What is the pH of the resulting buffer? HA (aq) + OH (aq) ---> H2O (L) + A (aq)

Answer 1

Answer – We are given, 500.0 mL of 0.140 M NaOH

595 mL of 0.200 M weak acid

Ka = 2.29*10^-5

First we need to calculate the moles of each

Moles of NaOH = 0.500 L * 0.140 M

                           = 0.070 moles

Moles of weak acid, HA = 0.595 L * 0.200 M

                                         = 0.119 moles

pKa = -log 2.29*10^-5

        = 4.64

HA (aq) + OH^- (aq) ---> H₂O (l) + A^- (aq)

0.119        0.070                              0.070 moles

Moles of after reaction

HA = 0.119 – 0.070 = 0.049 moles

A^- = 0.070 moles

Total volume = 500+595 = 1095 mL = 1.095 L

[HA] = 0.049 moles / 1.095 L = 0.0447 M

[A^-] = 0.070 moles / 1.095 L = 0.0639 M

Now using the Henderson-Hasselbalch equation

pH = pKa + log [A^-] /[HA]

      = 4.64 + log 0.0639 M / 0.0447 M

      = 4.8