Question

The volume of a sample of pure HCl gas was 277 mL at 20°C and 189 mmHg. It was completely dissolved in about 70 mL of water and titrated with an NaOH solution; 16.5 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

Answer 1

First calculate mole of HCl by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = atm pressure= 189 mmHg = 0.248684 atm,

V = volume in Liter = 277 ml = 0.277 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 20⁰C = 273.15+ 20 = 293.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.248684 0.277)/(0.08205 293.15) = 0.002865 mole of HCl

neutrilization reaction of HCl with NaOH is

HCl + NaOH NaCl + H₂O

thus 1 mole of HCl react with 1 mole of NaOH therefore to nutrilize 0.002865 mole of HCl require 0.002865 mole of NaOH

Hence mole of NaOH = 0.002865

molarity = no. of mole / volume of solution liter

molarity of NaOH = 0.002865/0.0165 L = 0.1736 M

molarity of NaOH =  0.1736 M