Question

A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 18.3 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample?

NaOH + HCl --> NaCl + H2O

Note:

1. Keep 3 sig figs for your final answer.

2. Answer as percentage.

Answer 1

Given the cncentration of HCl solution = 0.500 M

Volume of HCl required to reach the end point = 18.3 mL = 18.3 mL x (1L / 1000 mL) = 0.0183 L

Hence moles of HCl required to reach the end point = MxV = 0.500M x 0.0183L = 0.00915 mol

Let the mass of NaOH in the sample be 'm' g

Hence moles of NaOH in the sample = mass of NaOH / molecular mass of NaOH

= m g / 40.0 g/mol = m/40 mol

The neutralization reaction is

NaOH + HCl --> NaCl + H2O

1 mol, 1 mol-- 1 mol, 1 mol

In the above balanced reaction

moles of HCl = moles of NaOH

=>  0.00915 mol = m/40 mol

=> m = 0.00915 mol x 40 = 0.366 g

i.e mass of NaOH in the sample = 0.366 g

Hence percentage of NaOH in the sample = (0.366 g / 0.500 g) x 100 = 73.2 % (answer)