percent purity and titrations

Question

A 0.3270-g sample of impure Ca(OH)₂ is dissolved in enough water to make 47.20 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.1715-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 15.05 mL of the acid to reach the endpoint?

Answer 1

0.1715 M HCl * 0.01505 L = 2.58 * 10^-3 mol.

HCl reacts 1:2 with Ca(OH)2 by the equation:

2 HCl + Ca(OH)2 -> 2H2O + CaCl2

therefore 2.58 * 10^-3 mol HCl * 1 mol Ca(OH)2 / 2 mol HCl = 1.29 * 10^-3 mol Ca(OH)2

1.29 * 10^-3 mol Ca(OH)2 / 0.020 L = 0.0645 M Ca(OH)2

0.0645 mol / L* 0.04720 L * 74.09 g / mol = 0.2255 g Ca(OH)2

% purity = amount / sample * 100% -> 0.2256 / 0.3270 * 100% = 69% purity.

Hope this was useful to you, let me know if you have any questions!