Determine the pH of a \(0.50-M\) solution of \(\mathrm{NH}_{4} \mathrm{OCl.}\) . See Exercise \(181 .\) )

NH4OCl is a salt that is formed from the weak acid NH4+ and the weak base OCl-. When it dissolves in water, it will dissociate into those two ions. Thus, the dissociation reaction can be written as follows: NH4OCl (aq) → NH4+ (aq) + OCl- (aq)

Since NH4+ is a weak acid and OCl- is a weak base, we need to write K_a and K_b equilibrium expressions. For NH4+ (from NH4+ + H2O (l) ⇌ NH3 (aq) + H3O+(aq)): \[K_a = \frac{[\mathrm{NH}_3][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{NH}_4^+]}\] And for OCl- (from OCl- (aq) + H2O (l) ⇌ HOCl (aq) + OH-(aq)): \[K_b = \frac{[\mathrm{HOCl}][\mathrm{OH}^-]}{[\mathrm{OCl}^-]}\] We are given information about the concentration of NH4OCl, which is 0.50 M, and hence both NH4+ and OCl- will have an initial concentration of 0.50 M.

To determine the concentration of H3O+ and OH- at equilibrium, we can set up an equilibrium table for the weak acid and weak base reactions. Let x be the moles of NH4+ that dissociate into NH3, then the equilibrium concentrations are: NH4+: (0.50 - x) M NH3: x M H3O+: x M In a similar way, let y be the moles of OCl- that dissociate into HOCl, then the equilibrium concentrations are: OCl-: (0.50 - y) M HOCl: y M OH-: y M Now we can substitute these concentrations into the K_a and K_b expressions: \[K_a = \frac{x^2}{0.50-x}\] \[K_b = \frac{y^2}{0.50-y}\] The K_w value, which is equal to K_a * K_b, can be used to simplify the equation. K_w is equal to 1.0 × 10^-14 at 25°C. K_w = K_a * K_b = (x^2 / (0.50 - x)) * (y^2 / (0.50 - y))

First, we need to find the Ka and Kb values for the weak acid NH4+ and the weak base OCl- respectively from their respective parent acids and bases (NH4+ from NH3 and OCl- from HOCl with Ka given as 5.6 × 10^-10 and Kb given as 2.9 × 10^-7, respectively). We can solve for Ka and Kb from provided values, K_a = Kw / Kb and K_b = Kw / K_a. Then, we can use the values of Ka and Kb to solve for x and y, by solving the quadratic equations in the expressions. After we find x and y, we can find the pH by computing -log10(x), the negative base-10 logarithm of the H3O+ concentration. The PKa value of NH4+ and the pKb value of OCl- can also be calculated. If needed, you can also find that when x and y are very small, the concentration of NH4+ and OCl- remains 0.50 M approximately. Once you have these values, you can calculate the pH of the 0.50 M solution of NH4OCl by taking the -log10 of the concentration of H3O+.