Problem 68 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing \(0.25\) mole/L of ammonium chloride and \(0.05\) mole/ \(\mathrm{L}\) of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution: \(K_{b}\left[\mathrm{NH}_{4} \mathrm{OH}\right]=1.80 \times 10^{-5}\) \(K_{s p}\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]=6 \times 10^{-10}\) \(K_{s n}\left[\mathrm{Al}(\mathrm{OH})_{3}\right]=6 \times 10^{-32}\)

Step-by-Step Solution

Answer

[Mg^{2+}] \approx 4.6 \times 10^{-2} \, \text{mol/L}, [Al^{3+}] \approx 1.25 \times 10^{-8} \, \text{mol/L}

Step 1 - Calculate Hydroxide Ion Concentration

To find the hydroxide ion concentration \([OH^-]\) in the buffer solution, we use the Henderson-Hasselbalch equation. First, calculate the pOH using the equation: \[ pOH = pK_b + \log\left( \frac{[NH_4^+]}{[NH_4OH]} \right) \]Given factors are:\([NH_4^+] = 0.25\, \text{mol/L}\) and \([NH_4OH] = 0.05\, \text{mol/L}\), and \(pK_b = -\log(1.80 \times 10^{-5})\approx 4.74\). Thus,\[ pOH = 4.74 + \log\left( \frac{0.25}{0.05} \right) = 4.74 + 0.70 = 5.44 \] Now calculate \([OH^-]\):\[[OH^-] = 10^{-5.44} \approx 3.63 \times 10^{-6}\, \text{mol/L} \].

Step 2 - Find Concentration of Magnesium Ions

The solubility product \(K_{sp}\) for \(Mg(OH)_2\) is used to determine [Mg\(^{2+}\)]. The expression is:\[ K_{sp} = [Mg^{2+}][OH^-]^2 = 6 \times 10^{-10} \]Substitute \([OH^-]\) to find \([Mg^{2+}]\):\[ [Mg^{2+}] \cdot (3.63 \times 10^{-6})^2 = 6 \times 10^{-10} \]\[ [Mg^{2+}] = \frac{6 \times 10^{-10}}{(3.63 \times 10^{-6})^2} \approx 0.046 \times 10^{-10} = 4.6 \times 10^{-2} \text{ mol/L} \].

Step 3 - Find Concentration of Aluminium Ions

The solubility product \(K_{sn}\) for \(Al(OH)_3\) is used to determine \([Al^{3+}]\). The expression is:\[ K_{sn} = [Al^{3+}][OH^-]^3 = 6 \times 10^{-32} \]Substitute \([OH^-]\) to find \([Al^{3+}]\):\[ [Al^{3+}] \cdot (3.63 \times 10^{-6})^3 = 6 \times 10^{-32} \]\[ [Al^{3+}] = \frac{6 \times 10^{-32}}{(3.63 \times 10^{-6})^3} \approx 1.25 \times 10^{-8} \text{ mol/L} \].

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a convenient way to calculate the pH or pOH of a buffer solution. This equation provides a simple relationship amongst the dissociation constant and the concentrations of the acid and its conjugate base. For a base, the equation looks like this: \[ pOH = pK_b + \log\left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \] In our exercise, the acid is ammonium chloride \([NH_4^+]\), and the base is ammonium hydroxide \([NH_4OH]\). Given that the concentration of \([NH_4^+]\) is 0.25 mol/L and \([NH_4OH]\) is 0.05 mol/L, we can substitute these values into the equation. The pK_b is the negative logarithm of the base dissociation constant \( K_b \). In this case, \( K_b = 1.80 \times 10^{-5} \), so \( pK_b = -\log(1.80 \times 10^{-5}) \approx 4.74 \). Substituting all known values, we find: \[ pOH = 4.74 + \log\left( \frac{0.25}{0.05} \right) = 5.44 \] This allows us to determine the hydroxide ion concentration \([OH^-]\) by reversing the pOH to find \([OH^-]\) concentration. This technique becomes extremely useful when preparing buffer solutions in laboratory settings, allowing precise control over pH or pOH, as needed.

Solubility Product Constant

The solubility product constant \( (K_{sp}) \) helps us understand how much of a substance can dissolve in a solution before precipitation occurs. It's particularly useful when looking at slightly soluble salts such as magnesium hydroxide \(Mg(OH)_2\) and aluminum hydroxide \(Al(OH)_3\). The expression for \(K_{sp}\) takes the form of the product of the concentrations of the ions each raised to the power that corresponds to their stoichiometric coefficients in the balanced equation. For \(Mg(OH)_2\), given: - The solubility product is \(K_{sp} = 6 \times 10^{-10} \).- Its dissolution can be represented by \(Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-\). Thus, \[ K_{sp} = [Mg^{2+}][OH^-]^2 \] Similarly, for \(Al(OH)_3\): - The solubility product is \(K_{sn} = 6 \times 10^{-32} \).- Its dissolution can be represented by \(Al(OH)_3 \rightarrow Al^{3+} + 3OH^-\). Resulting in the equation: \[ K_{sn} = [Al^{3+}][OH^-]^3 \] These expressions allow chemists to compute the concentrations of ions that result from dissolving a given amount of a sparingly soluble compound in solution.

Ion Concentration Calculation

Calculating ion concentrations in buffer solutions involves using equilibrium constants like \(K_{sp}\) to predict how much of each ion will be present in the system. Let's go through it step by step. **Magnesium Ions \([Mg^{2+}]\)**: 1. Using the known \(K_{sp}\) for \(Mg(OH)_2\), which is \(6 \times 10^{-10}\).2. Substitute the near-calculated \([OH^-]\) concentration, which was \(3.63 \times 10^{-6} \text{ mol/L}\).3. Solve \([Mg^{2+}]\) by rearranging the equation: \[ [Mg^{2+}] \cdot (3.63 \times 10^{-6})^2 = 6 \times 10^{-10} \]4. Finding \([Mg^{2+}]\) gives approximately \(4.6 \times 10^{-2} \text{ mol/L}\). **Aluminium Ions \([Al^{3+}]\)**: 1. The \(K_{sn}\) for \(Al(OH)_3\) is much lower, \(6 \times 10^{-32}\).2. Again, substitute the same \([OH^-]\) value.3. Follow the same steps for \([Al^{3+}]\): \[ [Al^{3+}] \cdot (3.63 \times 10^{-6})^3 = 6 \times 10^{-32} \] 4. You'll find \([Al^{3+}]\) around \(1.25 \times 10^{-8} \text{ mol/L}\). Using such orderly calculation expressions helps predict the presence of specific metal ions in solution, which is fundamental to work involving reaction equilibria.

pH and pOH Concepts

Understanding the concepts of pH and pOH is critical in chemistry, especially when dealing with buffer solutions. These properties tell us about the acidity or basicity of a solution. - **pH**: This is a measure of the concentration of hydrogen ions \([H^+]\) in a solution. Lower pH values indicate higher acidity, and higher pH values suggest alkalinity. - **pOH**: This measures the hydroxide ion \([OH^-]\) concentration. It's a less common term but important in understanding basic solutions. Lower pOH values indicate higher alkalinity. The relationship between pH and pOH in aqueous solutions is governed by the equation: \[ pH + pOH = 14 \] (at 25°C) This means if one is known, the other can easily be deduced. In buffer solutions, the relationships get more specific, due to the buffering agents keeping the pH and pOH quite stable despite the addition of strong acids or bases. In our exercise, we calculated the pOH through the Henderson-Hasselbalch equation and fantastically saw how the buffer maintained the solution's stability. These concepts help in various applications like pharmaceuticals, biochemistry, and environmental science.