Question

An analyst is to determine the molarity of a NaOH solution. Three aliquots of standard KHP are weighed out: 0.3051 g, 0.4763 g, and 0.3803 g. These three aliquots were titrated using a NaOH solution and required 15.82 mL, 24.98 mL, and 18.65 mL, respectively. From these data calculate the average molarity of the NaOH solution.

Answer 1

Answer – First we need to calculate the moles of each given KHP.

Moles of KHP = 0.3051 g / 204.22 g.mol^-1 = 0.00149 moles

Moles of KHP = 0.4763 g / 204.22 g.mol^-1 = 0.00233 moles

Moles of KHP = 0.3803 g / 204.22 g.mol^-1 = 0.00186 moles

We know the KHP is monoprotic acid and NaOH is also monobasic, so

KHP + NaOH ------> H₂O + NaKP

So, 1 moles of KHP = 1 moles of NaOH

so , the moles of NaOH in the three aliquots are

0.00149 moles ,0.00233 moles and 0.00186 moles respectively and volume required 15.82 mL, 24.98 mL, and 18.65 mL, respectively

So molarity of NaOH = 0.00149 moles / 0.01582 L = 0.0944 M

molarity of NaOH = 0.00233 moles / 0.02498 L = 0.0933 M

molarity of NaOH = 0.00186 moles / 0.01865 L = 0.0998 M

so the average molarity of NaOH = 0.0944 + 0.0933 + 0.0998 / 3

                                                        = 0.0959 M