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Chapter 16: Problem 45

Calculate the solubility of solid $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\( in a \)0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}$ solution.

Short Answer

Expert verified
The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} M\).

Step by step solution

01

Write Balanced Chemical Equation

For the dissolution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in water, we can write the balanced chemical equation as follows: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca^{2+}}(aq) + 2\mathrm{PO^{3-}_{4}}(aq)\]
02

Write the Solubility Product Expression

Given the balanced chemical equation, we can now write the solubility product expression \(K_{sp}\) for the reaction. \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO^{3-}_{4}}]^2\]
03

Set Up the ICE Table

To find the equilibrium concentrations, we can set up an ICE (Initial, Change, Equilibrium) table for the reaction: \[\begin{array}{c|ccc} & \mathrm{Ca}^{2+} & \mathrm{PO}_{4}^{3-} \\ \hline \text{Initial} & 0 & 0.20 \\ \text{Change} & +3s & +2s \\ \text{Equilibrium} & 3s & 0.20 + 2s \end{array}\] Where s is solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).
04

Substituting the Equilibrium Concentrations into \(K_{sp}\) Value

Substituting the equilibrium concentrations into the solubility product expression, and using the given \(K_{sp} = 1.3 \times 10^{-32}\): \[1.3 \times 10^{-32} = (3s)^3(0.20 + 2s)^2\]
05

Solve for Solubility

Since \(K_{sp}\) is a very small number, we can approximate that \(0.20 + 2s \approx 0.20\). Therefore, the equation becomes: \[1.3 \times 10^{-32} = (3s)^3(0.20)^2\] Solve for s: \(s \approx 4.95 \times 10^{-11} \mathrm{M}\) The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} \mathrm{M}\).

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Most popular questions from this chapter

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(A g_{3} P O_{4}, K_{s p}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+} \right.$ is the cation in is the cation in solution.\()\)

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

The salt MX has a solubility of $3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\( in a solution with \)\mathrm{pH}=0.000 .\( If \)K_{\mathrm{a}}$ for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

Will a precipitate form when 100.0 \(\mathrm{mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to 100.0 \(\mathrm{mL}\) of \(2.0 \times 10^{-4} \mathrm{MNaOH}\)?

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)
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