Question

Titrations of 25.00 mL aliquots of CaOH₂(aq) solution (saturated at ~100 °C) with standardized 0.1020 moles∙L^-1 HCl(aq)

Final Burette Reading (mL)	10.29	19.56	31.92
Initial burette Reading (mL)	0.29	10.29	21.34
Volume of HCl(aq) added	10.00	9.27	9.33

volume of HCL 9.3mL

	Trial 1	Trial 2	Trial 3
Final Burette Reading (mL)	15.23	23.19	31.30
Initial burette Reading (mL)	8.41	16.61	23.91
Volume of HCl(aq) added	6.82	7.30	7.39

Volume of HCL 7.35mL

[OH^–] = __0.1789______________ mol L^-1

OH^–] = ___________0.0584_____ mol L^-1

1. Calculate the [Ca²⁺] in the saturated solution and report the molar solubility of Ca(OH)₂ (s) at both room and elevated temperatures.

Room temperature:

[Ca²⁺] = ________________ mol L^-1

s = ________________ mol L^-1

Elevated temperature:

[Ca²⁺] = ________________ mol L^-1

s = ________________ mol L^-1

Answer 1

The concentration of OH- and Ca + 2 in each trial is calculated, that of Ca + 2 is equal to the solubility:

[OH-] = Ca * Va / Vb = 0.1020 M * 10 mL / 25 mL = 0.0408 M

Trial 2: 0.0378 M

Trial 3: 0.0380 M

High temperature:

Trial 1: 0.0278 M

Trial 2: 0.0298 M

Trial 3: 0.0302 M

The concentration of Ca + 2 is calculated:

[Ca + 2] = [OH-] / 2 = 0.0408 M / 2 = 0.0204 M

Trial 2: 0.0189 M

Trial 3: 0.0190 M

High temperature:

Trial 1: 0.0139 M

Trial 2: 0.0149 M

Trial 3: 0.0151 M

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