Problem 82 - Chemistry: An Atoms-Focused Approach

Question

What is the pH of \(0.50 M \mathrm{CrCl}_{3} ?\)

Step-by-Step Solution

Answer

Answer: The pH of a 0.50 M CrCl3 solution is 6.45.

Step 1: Write the dissociation equation for CrCl3

CrCl3 is a salt that dissociates in water into its constituent ions. The dissociation equation is: $$\mathrm{CrCl}_{3(s)} \rightleftharpoons \mathrm{Cr^{3+}_{(aq)}} + 3\mathrm{Cl^-_{(aq)}}$$

Step 2: Note that \(\mathrm{Cr^{3+}}\) ions react with water to produce small concentration of \(\mathrm{H^+}\) ions

The Chromium(III) ion will hydrolyze with water to produce Hydrogen ions and the corresponding hydroxy complex ion. The hydrolysis reaction of Cr+3 ions with water is: $$\mathrm{Cr^{3+}_{(aq)}} + 6 \mathrm{H_2 O_{(l)}} \rightleftharpoons \mathrm{[Cr(H_2 O)_6]^{3+}_{(aq)}}$$ $$\mathrm{[Cr(H_2 O)_6]^{3+}_{(aq)}} \rightleftharpoons \mathrm{[Cr(H_2 O)_5 (OH)]^{2+}_{(aq)}} + \mathrm{H^+_{(aq)}}$$

Step 3: Calculate the equilibrium constant for the hydrolysis reaction of Cr+3 ion

The equilibrium constant \(K_\mathrm{a}\) for the reaction can be obtained from the literature. In this case, we can assume that the equilibrium constant \(K_\mathrm{a}\) is \(6.3 \times 10^{-13}\).

Step 4: Calculate the concentration of H+ ions

As we know the initial concentration of CrCl3 is 0.50 M, so we can assume the concentration of Cr+3 ions in solution is also 0.50 M. The equation for the hydrolysis reaction of Cr+3 ion can be represented as: $$\frac{[\mathrm{H^{+}}][\mathrm{[Cr(H_2 O)_5 (OH)]^{2+}}]}{[\mathrm{[Cr(H_2 O)_6]^{3+}}]} = K_\mathrm{a}$$ We can assume that the concentration of hydroxy complex [Cr(H2O)5(OH)]2+ is equal to the concentration of H+ ions formed, as both are produced in equal amounts according to the reaction. Therefore, we can write the equation as follows: $$\frac{[\mathrm{H^{+}}]^2}{[\mathrm{[Cr(H_2 O)_6]^{3+}}]} = 6.3 \times 10^{-13}$$ As initially, all Chromium ions are present in the form of [Cr(H2O)6]3+ (0.50 M), we can write the equation as follows: $$\frac{[\mathrm{H^{+}}]^2}{(0.50-[\mathrm{H^+}])} = 6.3 \times 10^{-13}$$ We can approximate as the concentration of H+ ions are very small compared to the initial concentration of Chromium ions, so the change in concentration can be ignored. Therefore: $$[\mathrm{H^{+}}]^2 = (0.50)(6.3 \times 10^{-13}),$$ thus $$[\mathrm{H^+}] = \sqrt{(0.50)(6.3 \times 10^{-13})} = 3.54 \times 10^{-7} \mathrm{M}$$

Step 5: Calculate the pH

Now we can find the pH using the definition of pH: $$\mathrm{pH} = -\log_{10} [\mathrm{H^+}]$$ $$\mathrm{pH} = -\log_{10} (3.54 \times 10^{-7}) = 6.45$$ So, the pH of the 0.50 M CrCl3 solution is 6.45.