Question

A quality control chemist at Dow Chemical tried to determine purity of NaOH using titration. He measured out 0.500 g NaOH sample and dissolved it in 20 mL water. 22.5 mL of a 0.500 mol/L HCl solution was used to reach the end point. Assume the impurity did not react with HCl, what is the % purity of this NaOH sample?

Answer 1

Answer:

89.94 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For HCl :

Molarity = 0.500 M

Volume = 22.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 22.5×10⁻³ L

Thus, moles of HCl :

$Moles=0.500 \times {22.5\times 10^{-3}}\ moles$

Moles of HCl = 0.01125 moles

The reaction of NaOH and HCl is:

NaOH + HCl ⇒ NaCl + H₂O

Moles of HCl = Moles of NaOH

Thus, Moles of NaOH = 0.01125 moles

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01125\ moles= \frac{Mass}{39.997\ g/mol}$

$Mass of NaOH= 0.4497\ g$

Total mass = 0.500 g

% purity = ( 0.4497 / 0.500 ) × 100 = 89.94 %