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Chapter 16: Problem 37

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 aluminum hydroxide \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(s = 10^{-8.5}\) mol/L.

Step by step solution

01

Write the chemical equation

Write the chemical equation for the dissolution of aluminum hydroxide \(\mathrm{Al}(\mathrm{OH})_{3}\) in water: \[\mathrm{Al}(\mathrm{OH})_{3} \rightleftharpoons \mathrm{Al}^{3+} + 3\mathrm{OH}^-\] This equation indicates that one mole of aluminum hydroxide dissociates into one mole of aluminum ions and three moles of hydroxide ions in aqueous solution.
02

Set up the \(K_{sp}\) expression

Write the \(K_{sp}\) expression in terms of the concentrations of aluminum and hydroxide ions: \[K_{sp} = [\mathrm{Al}^{3+}][\mathrm{OH}^-]^3\] The \(K_{sp}\) value is given as \(2 \times 10^{-32}\).
03

Define the molar solubility (s)

Let the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) be represented by the variable \(s\). At equilibrium, the molar concentrations of the ions can be represented as: \[[\mathrm{Al}^{3+}] = s\] \[[\mathrm{OH}^-] = 3s\]
04

Substitute the values into the \(K_{sp}\) expression

Substitute the expressions for the molar concentrations of the ions in the \(K_{sp}\) expression: \[\begin{aligned} 2 \times 10^{-32} &= (s)(3s)^3 \\ 2 \times 10^{-32} &= (s)(27s^3) \end{aligned}\]
05

Solve for s, the molar solubility

Now, solve for the molar solubility \(s\): \[\begin{aligned} s(27s^3) &= 2 \times 10^{-32} \\ 27s^4 &= 2 \times 10^{-32} \\ s^4 &= \frac{2 \times 10^{-32}}{27} \\ s^4 &= 7.41 \times 10^{-34} \end{aligned}\] Find the fourth root of the value to solve for \(s\): \[s = \sqrt[4]{7.41 \times 10^{-34}} = 10^{-8.5}\] Finally, we find the molar solubility of aluminum hydroxide \(\mathrm{Al}(\mathrm{OH})_{3}\) to be \(s = 10^{-8.5}\) mol/L.

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Most popular questions from this chapter

The \(K_{\mathrm{sp}}\) for silver sulfate $\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\( is \)1.2 \times 10^{-5} .$ Calculate the solubility of silver sulfate in each of the following. a. water b. \(0.10M\) \(\mathrm{AgNO}_{3}\) c. \(0.20M\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\)

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}\), $\mathrm{Na}_{2} \mathrm{S},\( and \)\mathrm{Na}_{3} \mathrm{PO}_{4} .$ What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are $K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},$ and $K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}.$

The copper(I) ion forms a chloride salt that has $K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\( Copper (I) also forms a complex ion with \)\mathrm{Cl}^{-} :$ $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\).

Silver chloride dissolves readily in \(2M \mathrm{NH}_{3}\) but is quite insoluble in \(2M \mathrm{NH}_{4} \mathrm{NO}_{3}\) . Explain.

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\( is \)6.8 \times 10^{-37}$ . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?
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