Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{Ksp}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(1.0 \times 10^{-15} . \mathrm{M}\). Find at what \(\mathrm{pH}\) does \(1.0 \times 10^{-3} . \mathrm{M} \mathrm{Al}^{3+}\) precipitate on the addition of buffer of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{4} \mathrm{OH}\) solution. (a) 10 (b) \(10.5\) (c) 11 (d) 12

Step-by-Step Solution

Answer

The pH at which precipitation occurs is 10.

Step 1: Understand the Reaction

The given problem involves the precipitation reaction of \(\text{Al(OH)}_3\), which dissociates as \[\text{Al(OH)}_3(s) \rightleftharpoons \text{Al}^{3+}(aq) + 3\text{OH}^-(aq)\]. The solubility product (\(K_{sp}\)) for this equilibrium is \(1.0 \times 10^{-15}\).

Step 2: Write the Equilibrium Expression

The equilibrium expression based on the dissolution of \(\text{Al(OH)}_3\) is given by \[K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3\].

Step 3: Substitute Known Values

We substitute the given values into the equation. Given \([\text{Al}^{3+}] = 1.0 \times 10^{-3} \, \text{M}\), the equation becomes \[1.0 \times 10^{-15} = (1.0 \times 10^{-3}) [\text{OH}^-]^3\].

Step 4: Solve for \([\text{OH}^-]\)

Rearrange the equation to solve for \([\text{OH}^-]\): \[[\text{OH}^-]^3 = \frac{1.0 \times 10^{-15}}{1.0 \times 10^{-3}}\]. Simplifying, \[[\text{OH}^-]^3 = 1.0 \times 10^{-12}\], so \([\text{OH}^-] = 10^{-4} \, \text{M}\).

Step 5: Convert \([\text{OH}^-]\) to pH

Using \([\text{OH}^-] = 10^{-4} \, \text{M}\), calculate pOH: \[\text{pOH} = -\log([\text{OH}^-]) = 4\]. Since \(\text{pH} + \text{pOH} = 14\), then \(\text{pH} = 14 - 4 = 10\).

Step 6: Conclusion

Hence, the \(\text{pH}\) at which \(\text{Al(OH)}_3\) begins to precipitate is 10.

Key Concepts

Solubility Product

The solubility product constant, often denoted as \( K_{sp} \), is a special type of equilibrium constant. It is used in chemistry to describe the saturation point of a sparingly soluble ionic compound. In simpler terms, \( K_{sp} \) tells us the maximum product of the concentrations of the constituent ions of a dissolved compound before the compound begins to precipitate out of solution. This is crucial for understanding when a solution will remain clear or when solids will begin to form.
For instance, with \( \text{Al(OH)}_3 \), knowing the \( K_{sp} \) value helps predict at what conditions the aluminum hydroxide will start to form a solid from the solution. Remember, a smaller \( K_{sp} \) value like \( 1.0 \times 10^{-15} \) indicates much lesser solubility, which means it will precipitate even with low ion concentrations.

Precipitation Equilibrium

Precipitation equilibrium involves a delicate balance between dissolution and precipitation reactions. When the product of the ion concentrations in a solution (known as the ionic product) equals the \( K_{sp} \), the solution is perfectly balanced at the point of saturation.
If the ionic product exceeds \( K_{sp} \), precipitation occurs as the ions combine to form a solid compound, as seen with aluminum hydroxide in the exercise. On the flip side, if the ionic product is less than \( K_{sp} \), the solution remains unsaturated, and no precipitation takes place.

• This equilibrium is pivotal in processes where the removal of ions by precipitation is desired, such as in purifying water or in chemical analysis.

pH Calculation

pH is a scale used to specify the acidity or basicity of an aqueous solution. In the context of the problem, the pH becomes a critical factor because hydronium and hydroxide ions are directly involved in the solubility product and precipitation processes.
The relation between hydrogen ions (\( [\text{H}^+] \)) and hydroxide ions (\( [\text{OH}^-] \)) is expressed by the water dissociation constant \( K_w \) such that:
\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \].
Understanding pH is vital because many precipitation reactions are sensitive to changes in acidity or basicity, influencing whether a compound will dissolve more or precipitate out.

Buffer Solution

A buffer solution is a solution that resists sudden changes in pH upon the addition of an acid or a base. This characteristic comes from its specific composition generally involving a weak acid or base and its corresponding conjugate salt.
In this exercise, a buffer made of \( \text{NH}_4\text{Cl} \) and \( \text{NH}_4\text{OH} \) is used to control the pH of the solution. The ability to maintain a relatively constant pH is critical, especially when controlling the conditions for the dissolution or precipitation of compounds like \( \text{Al(OH)}_3 \).

• By adding such a buffer, you can reduce the effect of external factors or unintended additions of acids/bases, thereby keeping the chemical reaction more predictable and stable.

Chemical Equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the concentrations of the reactants and products remain constant over time. This does not mean the reactions have stopped but that the forward and backward reactions occur at equal rates.
In solutions, equilibrium concepts apply dominantly in reactions like dissociation of \( \text{Al(OH)}_3 \), where after dissolution, ions can recombine to reach an equilibrium state; thus affecting solubility and banding together to form a precipitate.

• The notion of equilibrium constant \( K \), such as \( K_{sp} \), is vital for predicting the extent and direction of a reaction.
• Understanding these helps in controlling processes that depend on precipitation, dissolution, and reaction back-and-forth.