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 \(4-12.\)

Step by step solution

01

Reaction of Aluminum hydroxide in acidic solution

In acidic solutions, the excess of protons (H+) reacts with the hydroxide ion (OH-) present in Aluminum hydroxide: \[ Al(OH)_3(s) + 3H^+(aq) \rightleftharpoons Al^{3+}(aq) + 3H_2O(l) \]

02

Reaction of Aluminum hydroxide in basic solution

In basic solutions, the excess of hydroxide ions (OH-) reacts with Aluminum hydroxide to form the soluble complex ion: \[ Al(OH)_3(s) + OH^-(aq) \rightleftharpoons Al(OH)_4^-(aq) \] b) Deriving the solubility equation

03

Write the equilibrium expressions

Write the expressions for the solubility product \(K_\mathrm{sp}\) and the equilibrium constant \(K\): \[ K_\mathrm{sp} = [Al^{3+}][OH^-]^3 \] \[ K = \frac{[Al(OH)_4^-]}{[Al(OH)_3][OH^-]} \]

04

Relate the concentrations

Relate the concentrations of ions using the equilibrium constant for water \(K_\mathrm{w} = [H^+][OH^-]\): \[ [OH^-] = \frac{K_\mathrm{w}}{[H^+]} \]

05

Substitute and Rearrange

Substitute the relationships into the first equation and rearrange: \[ S = [Al^{3+}] + [Al(OH)_4^-] \] \[ S = \frac{K_\mathrm{sp}} {[OH^-]^3} + \frac{K[OH^-]}{[Al(OH)_3]} \] \[ S = \frac{(([H^+])^3K_\mathrm{sp})} {(K_\mathrm{w})^3} + \frac{((K_\mathrm{w})K)}{[H^+]} \] c) Plotting the solubility Using a graphing software or calculator, plot the function: \[ S = \frac{(([H^+])^3(2\times10^{-32}))} {(K_\mathrm{w})^3} + \frac{((K_\mathrm{w})40.0)}{[H^+]} \] Over the pH range 4-12: remember that the concentration of \([H^+]\) ion is given by \([H^+] = 10^{-\mathrm{pH}}\). The graph will show a distinct trend in solubility as the pH varies, displaying the solubility of aluminum hydroxide increasing as the acidity decreases or as the basicity increases.

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