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Chapter 15: Problem 33

Calculate the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2 \times 10^{-32}\).

Short Answer

Expert verified
The molar solubility of Al(OH)3 is approximately \(6.34 \times 10^{-9} \: M\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of Al(OH)3 in water

We start by writing the balanced chemical equation for the dissolution of Al(OH)3: \[ Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3OH^-(aq) \]
02

Define the equilibrium concentrations of the ions in terms of molar solubility

Let's denote the molar solubility of Al(OH)3 as \(x\). Since one mole of Al(OH)3 produces one mole of Al3+ and three moles of OH-, at equilibrium we have: \[ [Al^{3+}] = x \] \[ [OH^-] = 3x \]
03

Write the expression for the solubility product constant, Ksp

The expression for the solubility product constant is given by the product of the concentrations of the ions raised to their stoichiometric coefficients: \[ K_{sp} = [Al^{3+}] [OH^-]^3 \]
04

Substitute the equilibrium concentrations into the Ksp expression

Now we substitute the expressions for the equilibrium concentrations from Step 2 into the Ksp expression: \[ K_{sp} = (x)(3x)^3 \]
05

Solve for the molar solubility, x

Now we have an equation with only one variable, x. Plug in the value of \( K_{sp} = 2 \times 10^{-32} \), and solve for x: \[ 2 \times 10^{-32} = x (27x^3) \] \[ x = \sqrt[4]{\frac{2 \times 10^{-32}}{27}} \] \[ x = \sqrt[4]{7.41 \times 10^{-34}} \] Evaluating this expression, we get: \[ x \approx 6.34 \times 10^{-9} \: M \]
06

State the molar solubility

The molar solubility of Al(OH)3 is approximately \(6.34 \times 10^{-9} \: M\).

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Most popular questions from this chapter

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI} ?\)

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 $$ \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) $$

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

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, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) 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 \(\mathrm{pH}\) dependence of the solubility of Al(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 $$ S=\left[\mathbf{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right] $$ 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 \(4-12\).

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 \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).
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