Problem 112 - Chemistry and Chemical Reactivity

Question

Aluminum hydroxide reacts with phosphoric acid to give AIPO \(_{4} .\) The substance is used industrially in adhesives, binders, and cements. (a) Write the balanced equation for the preparation of AlPO \(_{4}\) from aluminum hydroxide and phosphoric acid. (b) If you begin with \(152 \mathrm{g}\) of aluminum hydroxide and \(3.00 \mathrm{L}\) of \(0.750 \mathrm{M}\) phosphoric acid, what is the theoretical yield of AlPO_? (c) If you place \(25.0 \mathrm{g}\) of \(\mathrm{AlPO}_{4}\) in \(1.00 \mathrm{L}\) of water, what are the concentrations of \(\mathrm{Al}^{3+}\) and \(\mathrm{PO}_{4}^{3-}\) at equilibrium? (Neglect hydrolysis of aqueous Al \(^{3+}\) and \(\mathrm{PO}_{4}^{3-}\) ions.) \(K_{\mathrm{sp}}\) for \(\mathrm{AlPO}_{4}\) is \(1.3 \times 10^{-20}\) (d) Does the solubility of AIPO increase ur decrease on adding HCl? Explain. (IMAGE CAN'T COPY)

Step-by-Step Solution

Answer

(a) \(\text{Al(OH)}_3 + \text{H}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + 3 \text{H}_2\text{O}\); (b) 238g AlPO_4; (c) \(1.14 \times 10^{-10} \text{ M of Al}^{3+} \) and \(\text{PO}_4^{3-} \); (d) Solubility increases.

Step 1: Write the balanced chemical equation

The reaction between aluminum hydroxide (Al(OH)_3) and phosphoric acid (H_3PO_4) produces aluminum phosphate (AlPO_4) and water. The unbalanced chemical equation is:\[ \text{Al(OH)}_3 + \text{H}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + \text{H}_2\text{O} \]Balancing the equation, we get:\[ \text{Al(OH)}_3 + \text{H}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + 3 \text{H}_2\text{O} \]

Step 2: Calculate moles and determine limiting reactant

First, calculate the moles of aluminum hydroxide and phosphoric acid. For aluminum hydroxide (molecular weight = 78 g/mol):\[ \text{Moles of Al(OH)}_3 = \frac{152 \, \text{g}}{78 \, \text{g/mol}} \approx 1.95 \, \text{mol} \]For phosphoric acid (Molarity = 0.750 M, Volume = 3.00 L):\[ \text{Moles of H}_3\text{PO}_4 = 0.750 \, \text{mol/L} \times 3.00 \, \text{L} = 2.25 \, \text{mol} \]Since the reaction equation is 1:1, subtract the moles of aluminum hydroxide and phosphoric acid to find the limiting reactant. Aluminum hydroxide is the limiting reactant because it has fewer moles.

Step 3: Calculate theoretical yield of AlPO4

Using the limiting reactant (aluminum hydroxide), calculate the theoretical yield of AlPO_4:The balanced equation shows that 1 mole of Al(OH)_3 produces 1 mole of AlPO_4. Since we have 1.95 mol of Al(OH)_3:\[ \text{Moles of AlPO}_4 = 1.95 \, \text{mol} \]The molecular weight of AlPO_4 is \(122 \text{ g/mol}\):\[ \text{Theoretical yield of AlPO}_4 = 1.95 \, \text{mol} \times 122 \, \text{g/mol} \approx 238 \text{ g} \]

Step 4: Calculate concentrations of ions in solution

For AlPO_4, we use the solubility product \(K_{sp}\) to find the concentrations of ions. The equilibrium equation is:\[ \text{AlPO}_4 (s) \rightleftharpoons \text{Al}^{3+} (aq) + \text{PO}_4^{3-} (aq) \]Given \(K_{sp} = 1.3 \times 10^{-20}\), and if \(s\) is the solubility in mol/L:\[ K_{sp} = [\text{Al}^{3+}][\text{PO}_4^{3-}] = s^2 = 1.3 \times 10^{-20} \]\[ s = \sqrt{1.3 \times 10^{-20}} \approx 1.14 \times 10^{-10} \text{ mol/L} \]Thus, both the concentrations of \(\text{Al}^{3+}\) and \(\text{PO}_4^{3-}\) are approximately \(1.14 \times 10^{-10} \text{ mol/L}\).

Step 5: Effect of HCl on solubility

Adding HCl increases the H+ concentration in the solution. This will react with PO_4^{3-}, forming HPO_4^{2-} or H_2PO_4^- and thus driving the equilibrium to the right, thereby increasing the solubility of AlPO_4.

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that determines the maximum amount of product that can be formed. It is completely consumed in the reaction, and once it is used up, no more product can be made. When we mix aluminum hydroxide (\(\text{Al(OH)}_3\)) and phosphoric acid (\(\text{H}_3\text{PO}_4\)), we need to find out which one is the limiting reactant.
To do this, we calculate moles for each reactant based on given mass or concentration: 1.95 moles of \(\text{Al(OH)}_3\) and 2.25 moles of \(\text{H}_3\text{PO}_4\). The balanced reaction is 1:1, so the reactant with fewer moles (\(\text{Al(OH)}_3\)) is the limiting reactant.
This means aluminum hydroxide will dictate how much aluminum phosphate (AlPO_4) can be produced, as it runs out first.

Theoretical Yield

The theoretical yield refers to the maximum amount of product that can be formed from a given amount of reactant under perfect conditions. It's calculated based on the limiting reactant. After identifying the limiting reactant (aluminum hydroxide) in our reaction, we find it can form 1.95 moles of AlPO_4.
To find the theoretical yield in grams, we use the molar mass of AlPO_4 which is 122 g/mol. Using the formula: \ \[\text{Theoretical Yield} = \text{moles of AlPO}_4 \times \text{molar mass of AlPO}_4\ \]we find: \ \[\text{Theoretical Yield} = 1.95 \text{ mol} \times 122 \text{ g/mol} \approx 238 \text{ g}\ \]
This provides the maximum product obtainable in ideal conditions.

Solubility Product

Solubility product (\(K_{sp}\)) is a constant for a specific substance at a given temperature. It represents the level at which a solute dissolves in solution. For AlPO_4, its \(K_{sp}\) is \(1.3 \times 10^{-20}\) at room temperature.
In equilibrium, AlPO_4 dissolves slightly in water to produce \(\text{Al}^{3+}\) and \( \text{PO}_4^{3-} \). The equation for solubility equilibrium is: \[ \text{AlPO}_4 (s) \rightleftharpoons \text{Al}^{3+} (aq) + \text{PO}_4^{3-} (aq)\]If \(s\) is the solubility, then \([\text{Al}^{3+}]=[\text{PO}_4^{3-}]=s\) and the \(K_{sp}\) expression is \(s^2 = 1.3 \times 10^{-20}\).
Solving for \(s\) gives: \(s = \sqrt{1.3 \times 10^{-20}} \approx 1.14 \times 10^{-10}\) M, indicating the extremely low solubility of AlPO_4 in water.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. In the case of AlPO_4 dissolving in water, the equilibrium is between the solid AlPO_4 and its dissolved ions \(\text{Al}^{3+}\) and \(\text{PO}_4^{3-}\).
At this point, the concentrations of reactants and products remain constant.
Introducing HCl affects this equilibrium. H+ ions from the acid combine with \(\text{PO}_4^{3-}\) ions to form \(\text{HPO}_4^{2-}\) or \(\text{H}_2\text{PO}_4^-\), decreasing \(\text{PO}_4^{3-}\) levels. This disturbance causes more AlPO_4 to dissolve to re-establish equilibrium, thus increasing its solubility.
This is an example of Le Chatelier’s Principle, which predicts that the system will adjust to reduce the impact of the imposed change.