A solution consisting of \(25.00 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) in \(178 \mathrm{~mL}\) of water is titrated with \(0.114 \mathrm{M}\) KOH. (a) How many \(\mathrm{mL}\) of \(\mathrm{KOH}\) are required to reach the equivalence point? (b) Calculate \(\left[\mathrm{Cl}^{-}\right],\left[\mathrm{K}^{+}\right],\left[\mathrm{NH}_{3}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) at the equivalence point. (Assume that volumes are additive.) (c) What is the \(\mathrm{pH}\) at the equivalence point?

The balanced equation for the titration of NH4Cl with KOH is: NH4Cl(aq) + KOH(aq) → NH3(aq) + KCl(aq) + H2O(l)

We are given 25.0 g of NH4Cl in the solution. To find the moles, divide the mass (25.0 g) by the molar mass of NH4Cl (53.49 g/mol): moles of NH4Cl = 25.0 g / 53.49 g/mol = 0.467 mol

From the balanced equation, we see that 1 mole of NH4Cl reacts with 1 mole of KOH. Therefore, the moles of KOH required to reach the equivalence point is the same as the moles of NH4Cl: 0.467 mol NH4Cl × (1 mol KOH / 1 mol NH4Cl) = 0.467 mol KOH

We are given a concentration of KOH at 0.114 M. To find the volume of KOH required, use the formula: Volume of KOH = moles of KOH / concentration of KOH Volume of KOH = 0.467 mol / 0.114 M = 4.10 L = 4100 mL

At the equivalence point, the moles of Cl- and K+ ions are equal to the moles of NH4Cl (0.467 mol). The total volume of the solution at the equivalence point is 178 mL of NH4Cl solution + 4100 mL of KOH solution = 4278 mL. Calculate the concentrations of Cl-, K+, and NH3 ions: [Cl-] = [K+] = 0.467 mol / 4.278 L = 0.109 M [NH3] is also equal to [K+]: [NH3] = 0.109 M At the equivalence point, the OH- ions come from the dissociation of NH3: NH3(aq) + H2O(l) <---> NH4+(aq) + OH-(aq) To find the concentration of OH- ions we need to calculate the equilibrium constant, Kb for this reaction using the equilibrium constant expression: Kb = [NH4+][OH-] / [NH3] Kb for NH3 is 1.8 x 10^{-5}. Since [NH4+] and [OH-] are equal, we can rewrite the equation as follows: Kb = [OH-]^2 / [NH3] Solving for [OH-]: [OH-]^2 = Kb[NH3] [OH-] = sqrt(Kb[NH3]) = sqrt(1.8 * 10^{-5} * 0.109) = 1.3 * 10^{-3} M

First, calculate the pOH = -log[OH-] = -log(1.3 * 10^{-3}) = 2.89. Now, use the relation pH = 14 - pOH: pH = 14 - 2.89 = 11.11 At the equivalence point, the pH of the solution is 11.11.