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Homework Statement
A 0.2600-g sample of impure Ca(OH)2 is dissolved in enough water to make 45.70 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2455-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 9.88 mL of the acid to reach the endpoint?
Homework Equations
The Attempt at a Solution
1) determine the number of mmol of HCl
20.00 mL*(.2455 mmol/1 mL) = 4.91 mmol
2) Convert mmol HCl to OH^1- using the notion that H^+ + OH^1- => H2O
4.9 mmol H^+ *(1mmol OH^1-/1 mmol H^1+) = 4.9 mmol OH^1-
3) Convert the mmol of OH^1- in 45.70 mL of the solution of Ca(OH)2
4.9 mmol OH^1- * (1mmol Ca(OH)2/1mmol (OH)^1-) *(74.1 mg Ca(OH)2/1mmol Ca(OH)2) = 363.09 mg Ca(OH)2
4.) I used the amount of Ca(OH)2 in 9.88 mL of solution of mass of Ca(OH)2
363.09 mg Ca(OH)2 * ((1000 mL/9.88 mL)) = 36750 mg = 36.75 g Ca(OH)2
5.) Use the mass of Ca(OH)2 and mass of pellets to now determine percent purity
36.75 g Ca(OH)2 /.2600 * 100 = 14134.61538 %
Shouldn't the percent purity be smaller than 100 %