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Chapter 16: Problem 35

Calculate the molar solubility of $\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-15}.$

Short Answer

Expert verified
The molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.44\times10^{-5}\) mol/L.

Step by step solution

01

Write the balanced dissolution reaction and corresponding equilibrium expression

The dissolution reaction for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is given by: \[ \mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(aq) + 2\mathrm{OH}^-(aq). \] The corresponding equilibrium expression for Ksp is given by: \[ K_{sp}=[\mathrm{Cd}^{2+}][\mathrm{OH}^-]^2, \] where \([\mathrm{Cd}^{2+}]\) and \([\mathrm{OH}^-]\) represent the equilibrium concentrations of cadmium ions and hydroxide ions, respectively.
02

Set up a reaction table

To determine the equilibrium concentrations, we need to set up a reaction table: | | \(\mathrm{Cd}^{2+}\) | \(\mathrm{OH}^-\) | |-------------|-------------|-----------| | Initial | 0 | 0 | | Change | +s | +2s | | Equilibrium | s | 2s | where s represents the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\). Now, substitute the equilibrium concentrations into the equilibrium expression: \[ K_{sp}=(s)(2s)^{2}. \]
03

Solve for the molar solubility

Plug the given Ksp value into the equation: \(5.9\times10^{-15}=(s)(2s)^{2}\). Solving for s: \[ 5.9\times10^{-15}=s(4s^2)\Rightarrow s^3=\frac{5.9\times10^{-15}}{4}, \] so: \[ s=\sqrt[3]{\frac{5.9\times10^{-15}}{4}} \approx 1.44\times10^{-5}. \] The molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.44\times10^{-5}\) mol/L.

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Most popular questions from this chapter

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. \(0.10M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) \(\mathrm{NaI}\)

Two different compounds have about the same molar solubility. Do they also have about the same \(K_{\text {sp}}\) value?

What mass of \(\mathrm{ZnS}\left(K_{\mathrm{sp}}=2.5 \times 10^{-22}\right)\) will dissolve in 300.0 \(\mathrm{mL}\) of \(0.050M\) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}.\)

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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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