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A group of students conducted a titration of 25.00-mL saturated Ca(OH)2 solution with 0.0480M HCl. The...

A group of students conducted a titration of 25.00-mL saturated Ca(OH)2 solution with 0.0480M HCl. The students found that 10.10-mL of acid was required to reach the equivalence point. Calculate the molar concentration of OH- and Ca2+, the molar solubility, and Ksp of the analyte. Show detailed calculation for each.

[OH-] in the analyte:

[Ca2+] in the analyte:

molar solubility of Ca(OH)2:

Ksp of Ca(OH)2:

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Answer #1

Molarity of acid = 0.0480 M

Volume of Acid used = 10.10 mL = 0.01010 L

Thus, Moles of acid required = Molarity * Volume = 0.0480 M* 0.01010 L = 0.0004848 moles

Reaction between HCl and Ca(OH)2 is:

2HCl + Ca(OH)2 --> CaCl2 + 2H2O

from reaction,

2 mol of acid reacts with 1 mol of Ca(OH)2

0.0004848 moles of acid reacts with 0.0002424 mole of Ca(OH)2

Volume of Ca(OH)2 = 25 mL = 0.025 L

Thus, Molarity of Ca(OH)2 = Moles/ Volume = 0.0002424 mole / 0.025 L = 0.009696 M

1 mol of Ca(OH)2 gives 1 mole of Ca2+ and 2 mol of OH-

So, moles of OH- = 2* 0.0002424 mole = 0.0004848 moles

Total volume = 25 + 10.10 = 35.10 mL =0.03510 L

[OH-] = moles/ Total volume = 0.0004848 moles/ 0.03510 L = 0.0138 M

[Ca2+] = 0.0002424/ 0.03510 = 0.0069 M

Molar solubility of Ca(OH)2 = 0.0069 M

Ca(OH)2 dissociates as:

Ca(OH)2 --> Ca2+ + 2OH-

Ksp = [Ca2+][OH-]2 = ( 0.0069 )(0.0138 )2 = 1.32 *10-6

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