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Chapter 16: Problem 99

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is $1.6 \times 10^{-7} \mathrm{mol} / \mathrm{L}\( at \)25^{\circ} \mathrm{C},$ calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Short Answer

Expert verified
The solubility product constant \(K_{sp}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is approximately \(1.47 \times 10^{-26}\).

Step by step solution

01

1. Write the dissolution equation

First, we write the equation for the dissolution of calcium phosphate in water: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\]
02

2. Define the solubility and solubility product

We are given the solubility of calcium phosphate, which is the concentration of the ions in a saturated solution at equilibrium. Let the solubility be represented by 's', which is given as \(1.6 \times 10^{-7} \mathrm{mol}/\mathrm{L}\). The solubility product \(K_{sp}\) is defined as the product of the concentrations of the ions in a saturated solution at equilibrium: \[K_{sp} = [\mathrm{Ca}^{2+}]^3 [\mathrm{PO}_{4}^{3-}]^2\]
03

3. Calculate ion concentrations

We see that for each mole of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) that dissolves, 3 moles of \(\mathrm{Ca}^{2+}\) and 2 moles of \(\mathrm{PO}_{4}^{3-}\) are formed. From the solubility, we can find the concentration of each ion: \[[\mathrm{Ca}^{2+}] = 3s = 3(1.6 \times 10^{-7}\,\mathrm{mol}/\mathrm{L}) = 4.8 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}\] \[[\mathrm{PO}_{4}^{3-}] = 2s = 2(1.6 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}) = 3.2 \times 10^{-7} \,\mathrm{mol}/\mathrm{L}\]
04

4. Calculate the solubility product \(K_{sp}\)

Now we can plug these concentrations back into the equation for the solubility product: \[K_{sp} = (4.8 \times 10^{-7})^3 (3.2 \times 10^{-7})^2\] With a calculator, we determine the value of \(K_{sp}\): \[K_{sp} = 1.47 \times 10^{-26}\] The solubility product constant \(K_{sp}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is approximately \(1.47 \times 10^{-26}\).

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Most popular questions from this chapter

A solution is prepared by mixing \(50.0 \mathrm{mL}\) of \(0.10M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{mL}\) of $1.0 \mathrm{M}\( \)\mathrm{KCl}$ . Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. $\left[K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_{2}(s) \text { is } 1.6 \times 10^{-5}.\right]$

The \(K_{\mathrm{sp}}\) for \(Q,\) a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is $[\mathrm{Xe}] 6 s^{1} 4 f^{4} 5 d^{10} .\( The \)\mathrm{X}^{-}$ anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{g}\) \(\mathrm{NaX}\) in \(150 .\) mL solution?

A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and $1.0 \times 10^{-3} M$ \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and $K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .$

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s),\) but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-}.\) In terms of solubility, All \((\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the pH dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3},\) as a function of \(\left[\mathrm{H}^{+}\right],\) obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is 40.0 and \(K_{\mathrm{sp}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\) Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the ph range \(4-12.\)
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