Question

An analyst is to determine the molarity of a NaOH solution. Three aliquots of standard KHP are weighed out: 0.3021g, 0.4859g, and 0.3806g. These three aliquots were titrated using a NaOH solution and required 15.10mL, 23.00mL, and 19.43mL, respectively. From these data calculate the average molarity of the NaOH solution.

Answer 1

Knowing molar mass of KHP, its molarity should be found in each case:

C₁=0.3021/204.2=0.0015(mol/L);

C₂=0.4859/204.2=0.0024(mol/L);

C₃=0.3806/204.2=0.0019(mol/L).

Now molality of NaOH may be found:

C₁=0.0015×1000/15.1=0.0993(mol/L);

C₂=0.0024×1000/23=0.1043(mol/L);

C₃=0.0019×1000/19.43=0.0978(mol/L).

Average molarity:

C=(C₁+C₂+C₃)/3=(0.0993+0.1043+0.0978)/3=0.1005(mol/L)~0.1(mol/L).