Question

A solution consisting of \(25.00 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) in \(178 \mathrm{~mL}\) of water is titrated with \(0.114 \mathrm{M} \mathrm{KOH}\). (a) How many milliliters 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?

Short answer

What are the concentrations of Cl⁻, K⁺, NH₃, and OH⁻ at the equivalence point? What is the pH at the equivalence point? The volume of 0.114 M KOH needed to react with 25.00 g of NH₄Cl is 4100 mL. At the equivalence point, the concentrations are as follows: [Cl⁻] = 0.1093 M, [K⁺] = 0.1093 M, [NH₃] = 0.1093 M, and [OH⁻] ≈ 0 M. The pH at the equivalence point is approximately 8.66.

Step-by-step solution

Calculate the moles of NH₄Cl in the solution

Given the mass of NH₄Cl is 25.00 g, we can find the moles using the molar mass of NH₄Cl, which is 53.5 g/mol. \( \mathrm{moles \ of \ NH}_{4}\mathrm{Cl} = \dfrac{25.00 \ g}{53.5 \ g/mol} = 0.4673 \ mol\)

Calculate the moles of KOH required to reach the equivalence point

At the equivalence point, the moles of KOH required is the same as the moles of NH₄Cl since the reaction is 1:1. moles of KOH = 0.4673 mol

Convert the moles of KOH to milliliters

We know the concentration of KOH is 0.114 M, so we can calculate the volume of KOH required: \(\mathrm{volume \ of \ KOH}=\dfrac{\mathrm{moles \ of \ KOH}}{\mathrm{concentration \ of \ KOH}}=\dfrac{0.4673 \ mol}{0.114 \ M}=4.100 \ L\) Then, convert to milliliters: \(\mathrm{volume \ of \ KOH} = 4.100 \ L \times 1000 \ \dfrac{mL}{L}=4100 \ mL\)

Determine the total volume of the solution at the equivalence point

At the equivalence point, the volumes of the initial solution and the added KOH are additive: \(\mathrm{total \ volume} = 178 \ mL + 4100 \ mL = 4278 \ mL\)

Calculate the concentrations of Cl⁻, K⁺, NH₃, and OH⁻

At the equivalence point, the moles of Cl⁻ from NH₄Cl, K⁺ from KOH, and NH₃ from reaction are all in the solution. Note that as the \(\mathrm{NH}_{4} \mathrm{Cl}\) reacts completely with \(\mathrm{KOH}\), the \(\mathrm{OH}^{-}\) concentration will be 0. Calculate the concentrations: \([\mathrm{Cl}^{-}] =\dfrac{0.4673\ mol}{4.278 \ L}=0.1093\ M\) \([\mathrm{K}^{+}] =\dfrac{0.4673\ mol}{4.278 \ L}=0.1093\ M\) \([\mathrm{NH}_{3}] =\dfrac{0.4673\ mol}{4.278 \ L}=0.1093\ M\)

Calculate the pH of the solution at the equivalence point

First, let's find the pOH using the equilibrium constant (Kb) for NH₃ (which is 1.8×10⁻⁵) and the equilibrium expression: \(Kb=\dfrac{[\mathrm{OH}^{-}][\mathrm{NH}_{4}^{+}]}{[\mathrm{NH}_{3}]}\) We can assume that [OH⁻] = [NH₄⁺] (x) and [NH₃] = 0.1093 - x at equilibrium. \(1.8×10^{-5}=\dfrac{x^2}{0.1093-x}\) Solving for x, we find that x ≈ 4.571×10⁻⁶, which represents the [OH⁻] concentration at equilibrium. Now, calculate the pOH: \(pOH= -\log{[\mathrm{OH}^{-}]}=-\log{(4.571×10^{-6})}\approx 5.34\) Finally, find the pH using the relationship between pH and pOH: \(pH = 14 - pOH = 14 - 5.34 = 8.66\) To summarize, a) The milliliters of KOH required to reach the equivalence point is 4100 mL. b) At the equivalence point, the concentrations are as follows: [Cl⁻] = 0.1093 M, [K⁺] = 0.1093 M, [NH₃] = 0.1093 M, and [OH⁻] ≈ 0 M. c) The pH at the equivalence point is approximately 8.66.