Problem 114 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Small amount of freshly precipitated \(\begin{array}{llll}\text { magnesium } & \text { hydroxides } & \text { are } & \text { stirred }\end{array}\) vigorously in a buffer solution containing \(0.25 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(0.05 \mathrm{M}\) of \(\mathrm{NH}_{4} \mathrm{OH}\). \(\left[\mathrm{Mg}^{2}\right]\) in the resulting solution is \(\left(K_{\mathrm{b}}\right.\) for \(\mathrm{NH}_{4} \mathrm{OH}=2.0 \times 10^{-5}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) \(\left.=8.0 \times 10^{-12}\right)\) (a) \(4 \times 10^{-6} \mathrm{M}\) (b) \(2 \times 10^{-6} \mathrm{M}\) (c) \(0.5 \mathrm{M}\) (d) \(2.0 \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of \(\text{[Mg}^{2+}\text{]}\) is closest to \(4 \times 10^{-6} M\), which is option (a).

Step 1 - Calculate the Hydroxide Ion Concentration

Set up an expression using the amphiprotic buffer equation that relates the conjugate acid and base concentrations and the base dissociation constant (Kb). For a buffer containing NH4OH (weak base) and NH4Cl (its salt), the equation is: \[ OH^- = \frac{K_b \times [NH_4OH]}{[NH_4Cl]} \] Substitute the given concentrations and Kb value into the buffer equation to find the concentration of hydroxide ions (OH-).

Step 2 - Solve for Hydroxide Ion Concentration

Using the values \(K_b = 2.0 \times 10^{-5}\), \([NH_4OH] = 0.05 M\), and \([NH_4Cl] = 0.25 M\), the hydroxide ion concentration is calculated as follows: \[ [OH^-] = \frac{2.0 \times 10^{-5} \times 0.05}{0.25} = 4.0 \times 10^{-6} M \]

Step 3 - Use Ksp to Determine \([Mg^{2+}]\) Concentration

Now that the hydroxide ion concentration is known, use the solubility product constant (Ksp) of Mg(OH)2 to find the magnesium ion concentration. The Ksp expression is: \[ K_{sp} = [Mg^{2+}] [OH^-]^2 \] Solve for \([Mg^{2+}]\) using the known Ksp and the hydroxide ion concentration.

Step 4 - Solve for Magnesium Ion Concentration

Substitute the Ksp value of \(8.0 \times 10^{-12}\) and the calculated hydroxide concentration into the Ksp expression to find \([Mg^{2+}]\): \[ [Mg^{2+}] = \frac{K_{sp}}{[OH^-]^2} = \frac{8.0 \times 10^{-12}}{(4.0 \times 10^{-6})^2} = 5.0 \times 10^{-6} M \] The concentration of magnesium ions is halfway between (a) and (b). Since Ksp is a constant, the concentration cannot be (c) or (d). Considering the significant figures, option (a) is closer to the calculated concentration.

Key Concepts

Understanding the Solubility Product Constant (Ksp)

Delving into physical chemistry, one critical concept is the solubility product constant, often abbreviated as Ksp, which provides insight into the solubility of sparingly soluble salts. Ksp is a special case of the equilibrium constant that applies to the dissolution of solids into their constituent ions in a saturated solution.

For a general salt, which dissociates into anions and cations, the solubility product is expressed as the product of the concentrations of the resulting ions, each raised to the power of its stoichiometric coefficient. In mathematical terms, for a salt AB that dissociates into A+ and B-, the Ksp would be written as: \[ K_{sp} = [A^+] [B^-] \]
In practice, this constant can help us predict whether a precipitate will form when solutions of two salts are mixed. If the ionic product of the concentrations exceeds the Ksp, precipitation occurs. Conversely, if the ionic product is less than the Ksp, the salt remains in solution.

In the example exercise, the Ksp of magnesium hydroxide, \(Mg(OH)_2\), determines the maximum concentration of magnesium ions, \([Mg^{2+}]\), that can exist in equilibrium with hydroxide ions, \([OH^-]\), in the buffer solution before precipitation begins.

Buffer Solution Calculations

Buffer solutions are marvelous tools in chemistry, maintaining a stable pH when small amounts of acid or base are added. These solutions typically consist of a weak acid or base and its salt—known as its conjugate counterpart. They function by neutralizing added acids or bases through equilibrium reactions.

To calculate the pH of a buffer solution, the Henderson-Hasselbalch equation is used for acidic buffers, whereas a similar approach is applied to basic buffers, considering the base dissociation constant (Kb). The underlying concept is that the pH (or pOH for basic buffers) depends on the ratio of the concentrations of the acid (or base) to its conjugate form. For a basic buffer like that in our example, composed of \(NH_4OH\) and \(NH_4Cl\), the equation is:\[ pH = pK_b + \text{log} \left( \frac{[Conjugate Base]}{[Conjugate Acid]} \right) \]
Understanding buffer calculations is crucial, as biological systems frequently rely on buffers to maintain homeostasis, and they are equally vital in many laboratory and industrial processes.

Conjugate Acid-Base Pairs

The dance of protons between species in solution is elegantly described by the concept of conjugate acid-base pairs. When an acid donates a proton, it becomes its conjugate base; similarly, when a base accepts a proton, it turns into its conjugate acid. This action is reversible and lies at the heart of acid-base chemistry, often depicted by the general reaction:\[ HA \rightleftharpoons A^- + H^+ \]
Where HA represents the acid, A- is its conjugate base, and H+ is the proton. In a solution, these pairs work together to resist changes in pH when small amounts of acids or bases are added, hence creating buffer solutions.

In the given exercise, \(NH_4OH\) and \(NH_4Cl\) form such a pair. \(NH_4OH\) acts as the weak base, and \(NH_4^+\) is its conjugate acid. By recognizing and utilizing these pairs, chemists can manipulate reactions, design buffer systems, and predict the behavior of substances in various chemical environments.