Question

a sample of impure potassium hydrogen phtalte (KHP,molar mass=204.22g) weighing 2.1283 g required 42.85ml of 0.1084 M sodium hydroxide solution for titration to the end point. calulate the percentage of KHP in the sample.

[KHP+NaOH->KNaP}+H2O

Answer 1

From Law of milliequivalence N1 V1 = N2 V2

Where N1 = Normality of NaOH = 0.1084N (Since Molarity = Normality for NaOH)

V1 = Volume NaOH = 42.85ml

N2 = noramlity of KHP

V2 = Volume of KHP

From Normality = (Weight / Equivalent Weight ) 1000 / V in ml

So the Above equation can be modified as N1 V1 = (weight of KHP/ Equivalent weight of KHP)1000

Equivalent weight of KHP = Its molecular weight ( EW = MW/No. of replaceable H⁺)

0.1084 * 42.85 = (Weight of KHP / 204.22)1000

Weight of KHP = (0.1084*42.85*204.22)/1000

                        = 0.94859 g

This much weight of KHP is reacted completely with NaOH out of 2.1283g

So purity of KHP % = (0.94859 / 2.1283)100

                               = 44.5703 %