Calculate the \(\mathrm{pH}\) of the following solutions: a. \(1.2 M \mathrm{CaBr}_{2}\) b. \(0.84 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}}\right.\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\) ) c. \(0.57 \mathrm{M} \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

Since CaBr2 is derived from a strong base (Ca(OH)2) and a strong acid (HBr), it will not hydrolyze in water, and the pH will be neutral. Therefore, the pH of the solution is 7.

Since C6H5NH3NO3 is derived from a weak base (C6H5NH2) and a strong acid (HNO3), we need to consider the hydrolysis of the salt and use the Kb of the weak base (3.8 × 10^{-10}) and the concentration of the salt (0.84 M) to find the pH. First, we'll write the hydrolysis equation: \(C_6 H_5 NH_3^+ + H_2 O \leftrightarrow C_6 H_5 NH_2 + H_3 O^+\) Now, create an ICE table for this reaction: ``` Initial 0.84 M 0 0 Change -x x x Equilibrium (0.84 - x) x x ``` Since Kb of C6H5NH2 is given, we can calculate the Ka for C6H5NH3+ using the relationship Ka × Kb = Kw. Kw is the ion product of water, which equals \(1 × 10^{-14}\). \(K_a = \frac{K_w}{K_b} = \frac{1.0 × 10^{-14}}{3.8 × 10^{-10}} = 2.63 × 10^{-5}\) Using the Ka expression, we can find x: \(K_a = \frac{x^2}{0.84-x} = 2.63 × 10^{-5}\) Assuming x << 0.84, then we have: \(x^2 ≈ 0.84 × 2.63 × 10^{-5}\) \(x ≈ 4.69 × 10^{-3}\) Since x represents the concentration of H3O+, the pH can now be calculated: \(\mathrm{pH} = -\log{(4.69 × 10^{-3})} ≈ 2.33\) So, the pH of the 0.84 M C6H5NH3NO3 solution is approximately 2.33.

Since KC7H5O2 is derived from a strong base (KOH) and a weak acid (HC7H5O2), we need to consider the hydrolysis of the salt and use the Ka of the weak acid (6.4 × 10^{-5}) and the concentration of the salt (0.57 M) to find the pH. First, we'll write the hydrolysis equation: \(C_7H_5O_2^- + H_2O \leftrightarrow HC_7H_5O_2 + OH^-\) Now, create an ICE table for this reaction: ``` Initial 0.57 M 0 0 Change -x x x Equilibrium (0.57 - x) x x ``` Using the Ka expression for the weak acid, we can find x: \(K_a = \frac{x^2}{0.57-x} = 6.4 × 10^{-5}\) Assuming x << 0.57, then we have: \(x^2 ≈ 0.57 × 6.4 × 10^{-5}\) \(x ≈ 3.00 × 10^{-3}\) Since x represents the concentration of OH-, we can find the pOH using: \(\mathrm{pOH} = -\log{(3.00 × 10^{-3})} ≈ 2.52\) Now we can calculate the pH using the relationship pH + pOH = 14: \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.52 ≈ 11.48\) So, the pH of the 0.57 M KC7H5O2 solution is approximately 11.48.