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Chapter 15: Problem 25

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} \mathrm{M} .\) Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{PbBr}_{2}\).

Short Answer

Expert verified
The solubility product constant (\(K_{\mathrm{sp}}\)) for \(\mathrm{PbBr}_{2}\) is \(3.89 \times 10^{-5}\).

Step by step solution

01

Write the dissolution equilibrium for \(\mathrm{PbBr}_{2}\)

The dissolution equilibrium for \(\mathrm{PbBr}_{2}\) can be written as follows: \[\mathrm{PbBr}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{Br}^-(aq)\]
02

Write the expression for \(K_{\mathrm{sp}}\) for \(\mathrm{PbBr}_{2}\)

Based on the equilibrium reaction, the \(K_{\mathrm{sp}}\) expression for \(\mathrm{PbBr}_{2}\) can be written as: \[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{Br}^-]^2\]
03

Calculate the concentration of \(\mathrm{Br}^-\) ions

Since the dissolution of one \(\mathrm{PbBr}_{2}\) unit generates 2 \(\mathrm{Br}^-\) ions, the concentration of \(\mathrm{Br}^-\) ions is twice that of \(\mathrm{Pb}^{2+}\) ions. Thus, we have: \[[\mathrm{Br}^-] = 2\times [\mathrm{Pb}^{2+}] = 2 \times 2.14 \times 10^{-2}\ \mathrm{M} = 4.28\times 10^{-2}\ \mathrm{M}\]
04

Substitute the concentrations into the \(K_{\mathrm{sp}}\) expression and solve

Now we will substitute the given concentration of \(\mathrm{Pb}^{2+}\) ions and the calculated concentration of \(\mathrm{Br}^-\) ions into the \(K_{\mathrm{sp}}\) expression: \[K_{\mathrm{sp}} = (2.14 \times 10^{-2})(4.28\times 10^{-2})^2\] \[K_{\mathrm{sp}} = 3.89 \times 10^{-5}\] Thus, the solubility product constant (\(K_{\mathrm{sp}}\)) for \(\mathrm{PbBr}_{2}\) is \(3.89 \times 10^{-5}\).

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Most popular questions from this chapter

$$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$

Consider a solution made by mixing \(500.0 \mathrm{mL}\) of \(4.0\) \(M\) \(\mathrm{NH}_{3}\) and \(500.0 \mathrm{mL}\) of \(0.40\) \(M\) \(\mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) & \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) & K_{1} &=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2} &=8.2 \times 10^{3} \end{aligned}$$ Determine the concentration of all species in solution.

The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3},\) having a molar mass of \(288 \mathrm{g} / \mathrm{mol},\) is \(3.60 \times 10^{-7} \mathrm{g} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) of the compound.

a. Calculate the molar solubility of AgI in pure water. \(K_{\mathrm{sp}}\) for AgI is \(1.5 \times 10^{-16}\) b. Calculate the molar solubility of AgI in 3.0 \(M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\). c. Compare the calculated solubilities from parts a and b. Explain any differences.

A solution is prepared by mixing \(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{mL}\) of \(8.0 M \mathrm{NaF}\). $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \quad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of \(\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+}\) \(\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},\) and \(\mathrm{BeF}_{4}^{2-}\) in this solution.
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