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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 A 50.00 mL sample of the clear saturated solution is withdrawn and requires \(10.7 \mathrm{mL}\) of \(0.1032 \mathrm{M} \mathrm{HCl}\) for its titration.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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Most popular questions from this chapter

What is the minimum \(\mathrm{pH}\) at which \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\) will precipitate from a solution that is \(0.086 \mathrm{M}\) in \(\mathrm{Cr}^{3+}(\mathrm{aq}) ?\)

An aqueous solution that \(2.00 \mathrm{M}\) in \(\mathrm{AgNO}_{3}\) is slowly added from a buret to an aqueous solution that is \(0.0100 \mathrm{M}\) in \(\mathrm{Cl}^{-}\) and \(0.250 \mathrm{M}\) in \(\mathrm{I}^{-}\). (a) Which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is the first to precipitate? (b) When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c) Is the separation of \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\) feasible by fractional precipitation in this solution?

Explain the important distinction between each pair of terms: (a) solubility and solubility product constant; (b) common-ion effect and salt effect; (c) ion pair and ion product.

To precipitate as \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s}),\) all the \(\mathrm{Ag}^{+}\) present in \(338 \mathrm{mL}\) of a saturated solution of \(\mathrm{AgBrO}_{3}\) requires \(30.4 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) measured at \(23^{\circ} \mathrm{C}\) and \(748 \mathrm{mm} \mathrm{Hg} .\) What is \(K_{\mathrm{sp}}\) for \(\mathrm{AgBrO}_{3} ?\)

Will \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) precipitate from a buffer solution that is \(0.45 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.35 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\) and also \(0.275 \mathrm{M}\) in \(\mathrm{Al}^{3+}(\mathrm{aq}) ?\) For \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=\) \(1.3 \times 10^{-33} ;\) for \(\mathrm{CH}_{3} \mathrm{COOH}, K_{\mathrm{a}}=1.8 \times 10^{-5}\).
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