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Chapter 15: Problem 42

Calculate the solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1 \times 10^{-54}\right)\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Short Answer

Expert verified
The solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution is approximately \(1.7 \times 10^{-18}\: M\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the dissociation of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Pb}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\]
02

Write the expression for the \(K_{sp}\)

The \(K_{sp}\) expression for the dissolution of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is: \[K_{sp} = [\mathrm{Pb}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\]
03

Set up the ICE table

Initial concentrations: \([\mathrm{Pb}^{2+}] = 0.10\: M\), \([\mathrm{PO}_{4}^{3-}] = 0\: M\) Changes: Assume \(s\) is the concentration of \(\mathrm{PO}_4^{3-}\) at equilibrium. Then, \([\mathrm{Pb}^{2+}]\) will increase by \(3s\) and \([\mathrm{PO}_{4}^{3-}]\) will increase by \(2s\) since 3 moles of \(\mathrm{Pb}^{2+}\) forms for every 2 moles of \(\mathrm{PO}_{4}^{3-}\). Equilibrium concentrations: \([\mathrm{Pb}^{2+}] = 0.10 + 3s\: M\), \([\mathrm{PO}_{4}^{3-}] = 2s\: M\)
04

Substitute the equilibrium concentrations into the \(K_{sp}\) expression

Now, we can plug the equilibrium concentrations back into the \(K_{sp}\) expression: \(K_{sp} = (0.10 + 3s)^3(2s)^2\) Given that \(K_{sp} = 1 \times 10^{-54}\), we have: \(1 \times 10^{-54} = (0.10 + 3s)^3(2s)^2\)
05

Solve for the solubility

To solve for the solubility, we can approximate \(𝑠<<0.10\) because \(K_{sp}\) is so small that it is reasonable to assume that \(𝑠\) has a very small value. Therefore, \(1 \times 10^{-54} = (0.10)^3(2s)^2\) Now, solve for \(s\): \(s = \frac{\sqrt{1 \times 10^{-54}}}{0.10^{\frac{3}{2}} \times 2}\) \(s \approx 1.7 \times 10^{-18}\) The solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution is approximately \(1.7 \times 10^{-18}\: M\).

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Most popular questions from this chapter

a. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\left.\mathrm{F}^{-} . \text {(For } \operatorname{Sr} \mathrm{F}_{2}, K_{\mathrm{sp}}=7.9 \times 10^{-10} .\right)\). b. Would the measured molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}} \text { for } \mathrm{HF} \text { is } 7.2 \times 10^{-4} .\right)\).

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10 M \operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) NaI

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

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{L}\) of \(3.0 M \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$ \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) $$
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