Question

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

Short answer

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

Step-by-step solution

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: \[ Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3OH^-(aq) \]

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: \[ [Al^{3+}] = x \] \[ [OH^-] = 3x \]

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: \[ K_{sp} = [Al^{3+}] [OH^-]^3 \]

Substitute the equilibrium concentrations into the Ksp expression

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

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: \[ x \approx 6.34 \times 10^{-9} \: M \]

State the molar solubility

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