Problem 5 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The solubility product of \(\mathrm{Cr}(\mathrm{OH})_{3}\) at \(298 \mathrm{~K}\) is \(6.0 \times 10^{-31}\). The concentration of hydroxide ions in a saturated solution of \(\mathrm{Cr}(\mathrm{OH})_{3}\) will be: (a) \(\left(2.22 \times 10^{-31}\right)^{1 / 4}\) (b) \(\left(18 \times 10^{-31}\right)^{1 / 4}\) (c) \(\left(18 \times 10^{-31}\right)^{1 / 2}\) (d) \(\left(4.86 \times 10^{-29}\right)^{1 / 4}\)

Step-by-Step Solution

Answer

Option (b) \(\left(18 \times 10^{-31}\right)^{1/4}\) is correct.

Step 1: Write the Dissociation Equation

The dissociation of chromium(III) hydroxide in water is given by the equation:\[\text{Cr(OH)}_3(s) \rightleftharpoons \text{Cr}^{3+}(aq) + 3\text{OH}^-(aq)\]

Step 2: Write the Solubility Product Expression

The solubility product constant (Ksp) expression for \(\text{Cr(OH)}_3\) is given by:\[K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3\]

Step 3: Set Up the Ksp Equation with Variables

Let the solubility of \(\text{Cr(OH)}_3\) be \(s\). Then- \([\text{Cr}^{3+}] = s\)- \([\text{OH}^-] = 3s\) Substitute these into the Ksp expression:\[6.0 \times 10^{-31} = s \cdot (3s)^3 = 27s^4\]

Step 4: Solve for Solubility (s)

Rearrange and solve for \(s\):\[27s^4 = 6.0 \times 10^{-31}\]Divide by 27:\[s^4 = \frac{6.0 \times 10^{-31}}{27}\]This simplifies to:\[s^4 = \frac{6.0}{27} \times 10^{-31} = 2.22 \times 10^{-32}\]Take the fourth root of both sides:\[s = \sqrt[4]{2.22 \times 10^{-32}}\]

Step 5: Calculate Hydroxide Ion Concentration

Since \([\text{OH}^-] = 3s\), substitute the value of \(s\) found in Step 4:\[[\text{OH}^-] = 3 \times \sqrt[4]{2.22 \times 10^{-32}}\]Since the answer choices are in a specific format, reassess the original Ksp equation:\[\left((18) \times 10^{-31}\right)^{1/4} = 3 \times \left(2.22 \times 10^{-32} \right)^{1/4}\]This corresponds to option (b).

Key Concepts

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance between a dissolved solute and its undissolved form. In the case of sparingly soluble salts like chromium(III) hydroxide, \(\text{Cr(OH)}_3\), the equilibrium is established when the rate at which the salt dissolves equals the rate at which it precipitates. This concept is crucial to understanding how much of a substance can dissolve in a solution before reaching saturation.
The equilibrium can be expressed using the solubility product constant, \(K_{sp}\), which indicates the extent to which a solid can dissolve in solution. A lower \(K_{sp}\) value implies a substance is less soluble. It's the product of the concentrations of the ions each raised to the power of their coefficients in the balanced chemical equation.
In aqueous solutions, we often explore solubility equilibrium through calculation of ion concentrations at this constant state. This helps us predict precipitation or the amount of ions that a solution can sustain before saturation.

Chromium Hydroxide Dissociation

The dissociation of chromium hydroxide in water is a process where \(\text{Cr(OH)}_3\) dissolves to yield \(\text{Cr}^{3+}\) ions and \(\text{OH}^-\) ions. This dissociation can be represented by the equation:
\[\text{Cr(OH)}_3(s) \rightleftharpoons \text{Cr}^{3+}(aq) + 3\text{OH}^-(aq)\]
Understanding this dissociation is key to addressing questions about solubility and related concepts. In the equation, one mole of \(\text{Cr(OH)}_3\) forms one mole of \(\text{Cr}^{3+}\) and three moles of \(\text{OH}^-\).
This stoichiometry is crucial when predicting concentrations of ions in solution. It informs the creation of the \(K_{sp}\) expression needed for solving equilibrium problems. The dissociation helps us comprehend how solids align with their ionic components in a saturated solution.

Hydroxide Ion Concentration

Hydroxide ion concentration in a saturated solution of a sparingly soluble salt like chromium hydroxide is a significant factor in solubility calculations. \(\text{OH}^-\) ions form in triplet amounts with each \(\text{Cr}^{3+}\) ion, based on the dissociation equation. This relationship makes \(\text{Cr(OH)}_3\) an excellent example of how multi-ion formation affects solubility.
In mathematical expressions, the concentration of hydroxide ions can be represented as \(3s\) where \(s\) is the solubility of \(\text{Cr(OH)}_3\). Knowing this allows us to express the hydroxide ion concentration as a multiple of the solubility, especially when setting up and solving \(K_{sp}\) equations.
In a laboratory or real-world chemical analysis, understanding how to determine this concentration is useful for predicting solubility changes, pH shifts, and potential precipitations. The concept also plays a significant role in complex equilibrium calculations involving multiple ion species.

Equilibrium Constant Calculations

Equilibrium constant calculations provide critical insight into solubility and precipitation reactions. For \(\text{Cr(OH)}_3\), the equilibrium constant, \(K_{sp}\), helps determine at what point the solid will saturate a solution. Knowing \(K_{sp}\) allows for calculation of ion concentrations in equilibrium.
The \(K_{sp}\) expression for \(\text{Cr(OH)}_3\) is represented as \([\text{Cr}^{3+}][\text{OH}^-]^3\). By substituting variables where \(s\) is the solubility, we can solve for ion concentrations: \(K_{sp} = s \cdot (3s)^3\).
Through these calculations, we can determine hydroxide ion concentration as a function of \(K_{sp}\). For example, dividing the given \(K_{sp}\) by coefficients derived from the balanced equation makes it possible to compute solubility electrochemically—ensuring our solutions remain accurate.
Grasping this concept is indispensable for predicting physicochemical behaviors in academic and industrial contexts. It's fundamental to determining ion concentration thresholds leading to saturation or precipitation.