Problem 43 - Chemistry: Principles and Reactions

Question

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

Step-by-Step Solution

Answer

Answer: The pH of the solution at the equivalence point is 11.11.

Step 1: Write the balanced chemical equation for the titration

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

Step 2: Calculate the moles of NH4Cl in the solution

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

Step 3: Use stoichiometry to determine the moles of KOH required to reach the equivalence point

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

Step 4: Calculate the volume of KOH solution required to reach the equivalence point

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

Step 5: Calculate the concentrations of Cl-, K+, NH3, and OH- ions at the equivalence point

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

Step 6: Calculate the pH of the solution at the equivalence point

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.

Key Concepts

Ammonium chloride

Ammonium chloride, or NH extsubscript{4}Cl, is a white crystalline salt commonly used in cleaning and personal care products. It's also an essential component in the titration process of this exercise. When dissolved in water, ammonium chloride dissociates into ammonium ions (NH extsubscript{4} extsuperscript{+}) and chloride ions (Cl extsuperscript{-}). This dissociation is crucial for the titration reaction with potassium hydroxide.

• Known for its solubility in water, making it ideal for aqueous solution preparation.
• Dissociation into ions enables various chemical reactions.

In our exercise, we started with 25.0 g of NH extsubscript{4}Cl. To find out how this interacts in the titration, we first convert this mass into moles, using the molar mass of NH extsubscript{4}Cl, which is 53.49 g/mol. This conversion allows us to use stoichiometric relationships to determine reactants and products throughout the titration process.

Potassium hydroxide

Potassium hydroxide (KOH) is a strong base used frequently in chemical titrations. It plays a key role in the neutralization of ammonium chloride to form various products. When KOH dissolves in water, it fully dissociates into potassium ions (K extsuperscript{+}) and hydroxide ions (OH extsuperscript{-}).

• This dissociation is important in titration for introducing OH extsuperscript{-} ions into the solution.
• The strength of KOH as a base helps in effectively reaching the equivalence point during titration.

In the given task, 0.467 mol of KOH is needed to react with the equivalent amount of NH extsubscript{4}Cl. With a concentration of 0.114 M, we calculate the volume of KOH required to accomplish this task, using the formula for volume: moles divided by molarity. The result is a fairly significant volume, highlighting the completeness of the reaction.

Equivalence point

The equivalence point in a titration is where the moles of the titrant equal the moles of the substance being titrated. In other words, all of the acid (or base) has completely reacted with the base (or acid) added. This is a critical part of the titration process, dictating when the titration should be stopped.

• Represents the completion of the reaction where utmost neutralization occurs.
• In our solution, the equivalence point is reached when all 0.467 mol of NH extsubscript{4}Cl has reacted with 0.467 mol of KOH.

Achieving the equivalence point ensures that subsequent calculations for ion concentrations and pH are accurate. After the equivalence point, the solution contains NH extsubscript{3}, KCl, and water, with NH extsubscript{3} dominating the equilibrium dynamics, altering the pH as we calculate next.

pH calculation

Calculating the pH at the equivalence point involves understanding the chemical equilibrium in the solution. After the neutralization, the solution predominantly contains NH extsubscript{3}, which acts as a weak base, affecting the pH.

• The presence of NH extsubscript{3} means that some OH extsuperscript{-} ions are formed from its interaction within the solution: NH extsubscript{3}(aq) + H extsubscript{2}O(l) ⇌ NH extsubscript{4} extsuperscript{+}(aq) + OH extsuperscript{-}(aq)
• To calculate [OH extsuperscript{-}], we use the base dissociation constant (K extsubscript{b}) for NH extsubscript{3}, which is 1.8 ⨉ 10 extsuperscript{-5}.

We calculate [OH extsuperscript{-}] = √(K extsubscript{b} × [NH extsubscript{3}]), leading to a pOH, which is then used to find the pH, with the relationship: pH = 14 - pOH. The resulting pH at the equivalence point is 11.11, slightly basic, characteristic of the weak base equilibrium in play.