Question

A sample containing both HCl and HNO3 had a density of 1.083 gram/mL. A 10.0 mL aliquot of this sample required 28.44 mL of 0.1163 M NaOH for titration. Another 10 mL aliquot of samples required 16.31 mL of 0.1216 M AgNO3 for titration.
Calculate the percent of HCl and percent of HNO3 in the sample.

Answer 1

AgNO3 reacts with HCl to give AgCl

AgNO3 + HCl ---> AgCl (s) + HNO3

AgNO3 moles = M x V ( in L) =0.1216 x 0.01631 = 0.002 = HCl moles

HCl mass = moles x molar mass = 0.002 x 36.46 g/mol = 0.0723 g

sample mass = volume x density = 10 ml x 1.083 g/ml = 10.83 g

NaOH moles reacted = M x V = 0.1163 x 0.02844 = 0.0033

HNO3 moles neutralised = 0.0033 ( since HNO3 and NaoH react in 1:1 ratio)

HNO3 mass = moles x molar mass = 0.0033 mol x 63 g/mol = 0.2084 g

Mass % of HCl = ( 100 x HCl mass) / ( sample mass) = 100 x 0.0723 / 10.083 = 0.72 %

Mass % of HNO3 = ( 100 x 0.2084 /10.083) = 2.07 %