A 0.2400-g sample of impure Ca(OH) 2 is dissolved inenough water to make 48.50 m

Question

A 0.2400-g sample of impure Ca(OH)2 is dissolved inenough water to make 48.50 mL of solution. 20.00 mL of theresulting solution is then titrated with 0.1819-M HCl. What is thepercent purity of the calcium hydroxide if the titration requires12.34 mL of the acid to reach the endpoint?

1 %

Explanation / Answer

Ca(OH)2 + 2HCl -> CaCl2 +2H2O n(Ca(OH)2) = m/M = 0.24/74.096 = 0.003239mol n(Ca(OH)2))titrated with HCl = 0.003239 x 20/48.50 =0.001336 m(Ca(OH)2 ) = 0.001336 x 74.096 = 0.09897g(theoretical) n(HCl) = cv = 0.1819 x 0.01234 = 0.002245mol n(Ca(OH)2) = 1/2 x 0.002245 = 0.001122mol m(Ca(OH)2)= 0.001122 x 74.096 = 0.08314g (actual) Percent purity = 0.08314/0.09897 x 100 = 84.00% Hope this helps!

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