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Chapter 15: Problem 38

Calculate the solubility of \(\operatorname{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a pH of \(11.00 .\)

Short Answer

Expert verified
The solubility of Cobalt(II) hydroxide, \(\mathrm{Co}\left(\mathrm{OH}\right)_{2}\), in the buffered solution with a pH of 11.00 is \(2.5\times10^{-10} M\).

Step by step solution

01

Write the dissociation equation and equilibrium expression

Cobalt(II) hydroxide dissociates in water as follows: \[ \mathrm{Co}\left(\mathrm{OH}\right)_{2}(s) \rightleftharpoons \mathrm{Co^{2+}}(aq) + 2\,\mathrm{OH^-}(aq) \] The solubility product constant, \(K_{sp}\), for this reaction can be written as: \[ K_{sp} = [\mathrm{Co^{2+}}] [\mathrm{OH^-}]^2 \] where \([\mathrm{Co^{2+}}]\) and \([\mathrm{OH^-}]\) are the equilibrium concentrations of the Cobalt(II) ion and the hydroxide ion in the solution, respectively. We are given the value of \(K_{sp}\) to be \(2.5\times10^{-16}\).
02

Determine the hydroxide ion concentration from the pH

Since the solution is buffered with a pH of 11.00, we can determine the hydroxide ion concentration using the relationship between pH, pOH, and the ion product of water (\(K_w\)): \[pH + pOH = 14\] First, we need to find the pOH: \[pOH = 14 - pH = 14 - 11.00 = 3\] Now, we can calculate the hydroxide ion concentration, \([\mathrm{OH^-}]\): \[[\mathrm{OH^-}] = 10^{-pOH} = 10^{-3} = 1\times10^{-3} M\]
03

Calculate the molar solubility using the hydroxide ion concentration

We'll now substitute the \([\mathrm{OH^-}]\) value into the equilibrium expression for the \(K_{sp}\) and solve for \([\mathrm{Co^{2+}}]\), which represents the molar solubility: \[K_{sp} = [\mathrm{Co^{2+}}] [\mathrm{OH^-}]^2\] \[2.5\times10^{-16} = [\mathrm{Co^{2+}}] (1\times10^{-3})^2\] Now, we solve for molar solubility, \([\mathrm{Co^{2+}}]\): \[ [\mathrm{Co^{2+}}] = \frac{2.5\times10^{-16}}{(1\times10^{-3})^2} = 2.5\times10^{-10}M\] Finally, we find that the solubility of Cobalt(II) hydroxide, \(\mathrm{Co}\left(\mathrm{OH}\right)_{2}\), in the buffered solution with a pH of 11.00 is \(2.5\times10^{-10} M\).

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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. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\mathrm{sp}}=5.2 \times 10^{-12}\) c. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-31}\)

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with HCl?What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10\) \(M\) \(\mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with \(\mathrm{HCl}\)?

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}:\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA},\) are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 \mathrm{M}\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00\).

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 \(\mathrm{PbBr}_{2}\).

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0\) \(M\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0\) \(M\) \(\mathrm{NH}_{3} ?\) $$\begin{array}{r} \mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \\ K=1.0 \times 10^{7} \\ \mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \\ K_{\mathrm{sp}}=5.9 \times 10^{-15} \end{array}$$
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