Titration of Ca(OH)2 with 0.0256 M HCL Volume of HCL added 32.80 mL mol of HCL=m

Question

Titration of Ca(OH)2 with 0.0256 M HCL Volume of HCL added 32.80 mL mol of HCL=mol OH- at 8.39x10-4 Sample of Ca(OH)2 used was 3.044 g Initial volume of Ca(OH)2 added was 25.00ml then diluted to 75ml What is the Molarity of OH- and Molar solubility of Ca(OH)2? Titration of Ca(OH)2 with 0.0256 M HCL Volume of HCL added 32.80 mL mol of HCL=mol OH- at 8.39x10-4 Sample of Ca(OH)2 used was 3.044 g Initial volume of Ca(OH)2 added was 25.00ml then diluted to 75ml What is the Molarity of OH- and Molar solubility of Ca(OH)2? Volume of HCL added 32.80 mL mol of HCL=mol OH- at 8.39x10-4 Sample of Ca(OH)2 used was 3.044 g Initial volume of Ca(OH)2 added was 25.00ml then diluted to 75ml What is the Molarity of OH- and Molar solubility of Ca(OH)2?

Explanation / Answer

mol of HCl used = MV = 32.80*0.0256 = 0.83968 mmol of HCl

mmol of OH- = mmol fo HCl = 0.83968

mmol of Ca(OH)2 = 1/2*mmol of OH- = 0.83968/2 = 0.41984 mmol

[Ca(OH)2] in undiluted system = (0.41984 /25) = 0.0167936 M

[Ca(OH)2] in diluted system = (0.41984 /75) = 0.005597 M

Molar solubility of Ca(OH)2 = 0.005597 M

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