Question

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4. A group of students conducted a titration of 25.00-ml saturated Ca(OH)2 solution with 0.0480M HCI. The students found that 10.10-mL of acid was required to reach the equivalence point. Calculate the molar concentration of OH and Ca, the molar solubility, and K. of the analyte. Show detailed calculation for each Detailed Method of Calculation (OH) in the analyte: Ca?) in the analyte: molar solubility of Ca(OH)2: Ksp of Ca(OH)2:

Answer 1

Concentration HCl = 0.0480 M

volume HCl = 10.10 mL

moles HCl = (concentration HCl) * (volume HCl)

moles HCl = (0.0480 M) * (10.10 mL)

moles HCl = 0.4848 mmol

moles OH^- in analyte = moles HCl

moles OH^- in analyte = 0.4848 mmol

[OH^-] = (moles OH^- in analyte) / (volume of saturated solution)

[OH^-] = (0.4848 mmol) / (25.00 mL)

[OH^-] = 0.01939 M

There is one ion of Ca²⁺ per two ions of OH^-

[Ca²⁺] = [OH^-] / 2

[Ca²⁺] = (0.01939 M) / 2

[Ca²⁺] = 0.009696 M

molar solubility of Ca(OH)₂ = [Ca²⁺] = [OH^-] / 2

molar solubility of Ca(OH)₂ = 0.009696 M

K_sp of Ca(OH)₂ = [Ca²⁺][OH^-]²

K_sp of Ca(OH)₂ = (0.009696 M) * (0.01939 M)²

K_sp of Ca(OH)₂ = 3.646 x 10^-6