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Chapter 18: Problem 14

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) is shaken with water to produce a saturated solution. A 50.00 mL sample of the clear saturated solution is withdrawn and requires \(10.7 \mathrm{mL}\) of \(0.1032 \mathrm{M} \mathrm{HCl}\) for its titration. What is \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} ?\)

Short Answer

Expert verified
The solubility product constant for \(Ca(OH)_{2}\) is \(0.00535\).

Step by step solution

01

Identify Reaction

Write out the chemical equation of \(\mathrm{Ca(OH)_{2}}\) dissolving in water and the neutralisation with \(HCl\):\[\mathrm{Ca(OH)_{2} \rightleftharpoons Ca^{2+} + 2OH^{-}}\]\[\mathrm{OH- + H+ \rightarrow H2O}\]
02

Molar Concentration of \(HCl\)

Calculate moles of \(HCl\), which is equal to its molarity multiplied by its volume (converted to liters). Here, \(HCl\) moles = molarity * volume = \(0.1032 M * 0.0107 L = 0.00110344 mol\)
03

Finding Hydroxide Ions Concentration

The stoichiometric ratio between \(HCl\) and \(OH^-\) is 1:1, hence moles of \(OH^-\) = moles of \(HCl\) = 0.00110344 mol. The molar concentration of \(OH^-\) will then be its moles divided by the volume of the solution. So, \([OH^-] = \frac{0.00110344 mol}{0.05 L} = 0.0220688 M\)
04

Finding Calcium Ions Concentration

The stoichiometric ratio of \(Ca^{2+}\) to \(OH^-\) is 1:2. Hence, \([Ca^{2+}] = \frac{1}{2} [OH^-] = \frac{1}{2} * 0.0220688 = 0.0110344 M\)
05

Find Solubility Product Constant (\(K_{sp}\))

The solubility product (\(K_{sp}\)) is obtained by taking the product of the concentrations each raised to the power of the coefficient in the solubility reaction. This means, \(K_{sp} = [Ca^{2+}\)][OH^-]^2 = \(0.0110344(0.0220688)^2 = 0.00535\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Saturated Solution
A saturated solution occurs when a solid solute is in equilibrium with its dissolved ions in a solvent. In this context, when excess calcium hydroxide, \(\mathrm{Ca(OH)_2}\), is mixed with water, it dissolves until the solution can dissolve no more. At this point, the concentration of the dissolved ions \(\mathrm{Ca^{2+}}\) and \(\mathrm{OH^-}\) remains constant, even if more solid is added. This equilibrium means the solution is 'saturated.'

Key characteristics of a saturated solution:
  • Contains the maximum concentration of ions under specific conditions.
  • Additional solute will not dissolve unless conditions change (e.g., temperature).
  • An example of equilibrium in a chemical system: \(\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-}\).
Understanding this concept helps in calculating properties like the solubility product constant, which we explore next.
Neutralization Titration
Neutralization titration is a method used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. In the example provided, \(\mathrm{HCl}\) is used to titrate the saturated solution of \(\mathrm{Ca(OH)_2}\).

Steps involved in neutralization titration:
  • A known volume of saturated solution is taken (50.00 mL in this case).
  • A measured volume of \(\mathrm{HCl}\) is added until the reaction with \(\mathrm{OH^-}\) ions is complete, i.e., the endpoint.
  • The amount of \(\mathrm{HCl}\) needed helps calculate the \(\mathrm{OH^-}\) concentration.
This approach involves reactions like \(\mathrm{OH^- + H^+ \rightarrow H_2O}\), demonstrating the stoichiometry between the titrant and solution.
Solubility Product Constant
The solubility product constant, \(K_{sp}\), provides insight into the solubility of a compound in a solution. It is specific for sparingly soluble compounds like \(\mathrm{Ca(OH)_2}\).

To find \(K_{sp}\):
  • Determine the concentrations of ions at saturation: [\(\mathrm{Ca^{2+}}\)] and [\(\mathrm{OH^-}\)].
  • Use the equilibrium equation: \(K_{sp} = [Ca^{2+}][OH^-]^2\).
  • Plug in the calculated concentrations: \(0.0110344 \, M\) for \([Ca^{2+}]\) and \(0.0220688 \, M\) for \([OH^-]\).
  • Calculate: \(K_{sp} = 0.0110344 \, (0.0220688)^2 = 0.00535\).
This constant helps predict how much of the compound can dissolve in water, influencing various applications in chemistry.

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Most popular questions from this chapter

Which of the following solids are likely to be more soluble in acidic solution and which in basic solution? Which are likely to have a solubility that is independent of pH? Explain. (a) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} ;\) (b) \(\mathrm{MgCO}_{3} ;\) (c) \(\mathrm{CdS}\); (d) \(\mathrm{KCl} ;\) (e) \(\mathrm{NaNO}_{3} ;\) (f) \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Appendix E describes a useful study aid known as concept mapping. Using the methods presented in Appendix \(\mathrm{E},\) construct a concept map that links the various factors affecting the solubility of slightly soluble solutes.

Which of the following solids is (are) more soluble in a basic solution than in pure water: \(\mathrm{BaSO}_{4}, \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), \(\mathrm{Fe}(\mathrm{OH})_{3}, \mathrm{NaNO}_{3},\) or MnS? Explain.

Fluoridated drinking water contains about 1 part per million (ppm) of \(\mathrm{F}^{-}\). Is \(\mathrm{CaF}_{2}\) sufficiently soluble in water to be used as the source of fluoride ion for the fluoridation of drinking water? Explain. [Hint: Think of 1 ppm as signifying \(1 \mathrm{g} \mathrm{F}^{-}\) per \(10^{6} \mathrm{g}\) solution.

Which of the following has the highest molar solubility? (a) \(\mathrm{MgF}_{2}, K_{\mathrm{sp}}=3.7 \times 10^{-8}\) \(\mathrm{MgCO}_{3}\), \(K_{\mathrm{sp}}=3.5 \times 10^{-8} ;(\mathrm{c}) \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-25}\); (d) \(\mathrm{Li}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=3.2 \times 10^{-9}\).
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