Question

A 0.8126 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL of solution. It required 22.07 mL of NaOH to reach the endpoint in the titration. What is the molarity of the NaOH solution?

Answer 1

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 0.8126 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(0.8126 g)/(36.46 g/mol)

= 2.229*10^-2 mol

According to balanced equation

mol of NaOH reacted = (1/1)* moles of HCl

= (1/1)*2.229*10^-2

= 2.229*10^-2 mol

This is number of moles of NaOH

volume , V = 22.07 mL

= 2.207*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 2.229*10^-2/2.207*10^-2

= 1.01 M

Answer: 1.01 M