Question

A 0.4308g sample of KHP required 21.31mL of NaOH to reach a pink phenolphthalein endpoint. A second sample contained 0.3711g of KHP and required 17.99mL of NaOH to react. What is the average molarity of the NaOH? Also comment on the precision of the two trials.

Answer 1

molar Mass of KHP = 204.22 g / mol

=> Moles of KHP = 0.4308 / 204.22 = 2.11 x 10^-3 moles

At endpoint,

Moles of KHP = Moles of NaOH

=> Moles of NaOH = 2.11 x 10^-3

Moles = Molarity x Volume (L)

=> 2.11 x 10^-3 = Molarity x 0.02131

=> Molarity = 0.099 M

Sample 2

Moles of KHP = 0.3711 / 204.22 = 1.817 x 10^-3

=> Mmoles of NaOH = 1.817 x 10^-3

=> Molarity = 1.817 x 10^-3 / 0.01799 = 0.101 M

Average = (0.099 + 0.101) / 2 = 0.1 M

Since the value of molarity from the two samples are nearly same, the trials were quite precise