Problem 276 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(25^{\circ} \mathrm{C}\), the solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.0 \times 10^{-11} .\) At which \(\mathrm{pH}\), will \(\mathrm{Mg}^{2+}\) ions start precipitating in the form of \(\mathrm{Mg}(\mathrm{OH})_{2}\) from a solution of \(0.001\) M \(\mathrm{Mg}^{2+}\) ions? (a) 9 (b) 10 (c) 11 (d) 8

Step-by-Step Solution

Answer

(b) 10

Step 1: Understanding the Reaction

The dissolution of \(\mathrm{Mg}(\mathrm{OH})_2\) in water can be described by the equation: \(\mathrm{Mg}(\mathrm{OH})_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^-(aq)\). This tells us that for every mole of \(\mathrm{Mg}(\mathrm{OH})_2\) that dissolves, it forms one mole of \(\mathrm{Mg}^{2+}\) ions and two moles of \(\mathrm{OH}^-\) ions.

Step 2: Expressing the Solubility Product

The solubility product expression for \(\mathrm{Mg}(\mathrm{OH})_2\) is given by \( K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^-]^2 \). With a given \(K_{sp}\) value of \(1.0 \times 10^{-11}\), we can determine at what \(\mathrm{OH}^-\) concentration precipitation will occur.

Step 3: Determine OH- Concentration

Rearrange the solubility product equation as \([\mathrm{OH}^-]^2 = \frac{K_{sp}}{[\mathrm{Mg}^{2+}]}\). Substitute the given values to find \([\mathrm{OH}^-]\): \([\mathrm{OH}^-]^2 = \frac{1.0 \times 10^{-11}}{0.001} = 1.0 \times 10^{-8}\). Thus, \([\mathrm{OH}^-] = 1.0 \times 10^{-4}\).

Step 4: Relate OH- Concentration to pH

The \(\mathrm{pOH}\) of a solution is given by \(\mathrm{pOH} = -\log([\mathrm{OH}^-])\). Calculate \(\mathrm{pOH}\) using \([\mathrm{OH}^-] = 1.0 \times 10^{-4}\): \(\mathrm{pOH} = -\log(1.0 \times 10^{-4}) = 4\).

Step 5: Calculate pH from pOH

Knowing that \(\mathrm{pH} + \mathrm{pOH} = 14\) for a solution at \(25^{\circ}\mathrm{C}\), we can find \(\mathrm{pH}\): \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4 = 10\).

Step 6: Conclusion

Thus, \(\mathrm{Mg}^{2+}\) ions will start to precipitate as \(\mathrm{Mg}(\mathrm{OH})_2\) at a \(\mathrm{pH}\) of 10. This corresponds to answer choice (b) 10.

Key Concepts

Magnesium Hydroxide

Magnesium hydroxide, often encountered in chemistry, is a compound that plays a significant role in both laboratory and industrial processes. It's chemical formula is \( \mathrm{Mg(OH)}_2 \), and it is characterized as a slightly soluble ionic compound. Because it dissolves sparingly in water, it's the perfect example for demonstrating concepts related to solubility and precipitation.
When magnesium hydroxide dissolves in water, it does so in a reversible reaction:

• \( \mathrm{Mg(OH)}_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^-(aq) \)

This equation highlights that one mole of magnesium hydroxide forms one mole of magnesium ions (\( \mathrm{Mg}^{2+} \)) and two moles of hydroxide ions (\( \mathrm{OH}^- \)).
Understanding this ratio is crucial when calculating the solubility product or predicting the onset of precipitation.

pH Calculation

The concept of pH is integral when working with solutions. pH is a measure of the hydrogen ion concentration in a solution, but it's equally essential when dealing with hydroxide ions due to the relationship between pH and pOH.
To calculate the pH of a solution, it's often necessary to first determine the concentration of hydroxide ions. After finding \( \mathrm{OH}^- \), the pOH can be calculated using:

• \( \mathrm{pOH} = -\log([\mathrm{OH}^-]) \)

By exploiting the relationship between pH and pOH:

• \( \mathrm{pH} + \mathrm{pOH} = 14 \)

at room temperature (\(25^{\circ} \mathrm{C} \)), one can find the pH easily. This relationship is vital in predicting when compounds like magnesium hydroxide will start precipitating.

Precipitation

Precipitation occurs when the conditions of a solution allow a solute to transition from an aqueous state to a solid form. This process is fundamentally connected to the solubility product, \( K_{sp} \), of a compound.
Magnesium hydroxide, for example, precipitates from a solution when the concentration product of its ions exceeds the solubility product constant, \( K_{sp} \). In simpler terms, as the solution's pH values increase, the concentration of \( \mathrm{OH}^- \) increases, leading to precipitation when the ionic product \( [\mathrm{Mg}^{2+}][\mathrm{OH}^-]^2 \) surpasses \( 1.0 \times 10^{-11} \).
Observing precipitation helps chemists determine critical points of saturation and how acid-base equilibrium influences solubility.

Ksp Expression

The solubility product constant, \( K_{sp} \), is a measure of the solubility of ionic compounds in water. It's specific to each compound and indicates the equilibrium point of the dissolution process.
For magnesium hydroxide, \( K_{sp} \) is expressed as:

• \( K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^-]^2 \)

This expression signifies how the concentrations of dissolved ions relate to the point of saturation in the solution.Additionally, \( K_{sp} \) is crucial for determining the point at which precipitation will begin. Understanding and employing the \( K_{sp} \) value allows chemists to predict and control chemical reactions that involve sparingly soluble compounds.

Chemical Equilibria

Chemical equilibria describe the state where the concentrations of reactants and products remain constant over time in a closed system. This principle underlies and maintains the balance between dissolved ions and their compounds in water.
In the case of magnesium hydroxide's dissolution:

• The equilibrium \( \mathrm{Mg(OH)}_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^-(aq) \)

indicates a dynamic state where the rate of dissolving equals that of precipitation. At chemical equilibrium, any changes such as a shift in pH or concentration can influence the balance.
Studying chemical equilibria provides insights into reaction dynamics and the conditions required to manipulate solubility and precipitation behavior. It is a cornerstone concept in understanding how solutions react to different factors like temperature and concentration changes.