Question

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.030 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 \mathrm{M} \mathrm{HCl}\). Calculate the \(\mathrm{pH}\) after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL}\), (b) \(10.0 \mathrm{~mL}_{\text {, }}\) (c) \(20.0 \mathrm{~mL}\), (d) \(35.0 \mathrm{~mL}\), (e) \(36.0 \mathrm{~mL}\), (f) \(37.0 \mathrm{~mL}\).

Short answer

(a) pH = 11.34 (b) pH = 10.07 (c) pH = 9.25 (d) pH = 5.96 (e) pH = 5.65 (f) pH = 5.35

Step-by-step solution

Write the balanced chemical equation for the reaction

The reaction between ammonia (NH3) and hydrochloric acid (HCl) can be represented by the following balanced chemical equation: NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq)

Determine the initial concentration of NH3 and HCl

Initially, we have 30.0 mL of 0.030 M NH3 solution and 0 mL of 0.025 M HCl.

Calculate the moles of NH3 and HCl at each given point of the titration

(a) 0 mL HCl: No reaction has occurred, so NH3 remains 0.030 mol/L (b) 10.0 mL HCl: moles of HCl = (0.025 mol/L) × (10.0 mL) × (1 L/1000 mL) = 0.00025 mol (c) 20.0 mL HCl: Same calculation as in (b), but with 20.0 mL (d) 35.0 mL HCl: Same calculation as in (b), but with 35.0 mL (e) 36.0 mL HCl: Same calculation as in (b), but with 36.0 mL (f) 37.0 mL HCl: Same calculation as in (b), but with 37.0 mL

Calculate how much reactant is left or produced at each point in titration

(a) 0 mL HCl: No reaction has occurred, so [NH3] = 0.030 mol/L and [NH4+] = 0 (b) to (f): - Calculate the moles of NH3 remaining based on the moles of HCl added. - Determine the change in moles of NH4+ by comparing with the initial moles of NH3.

Calculate the pH at each point in titration

(a) pH from initial NH3 concentration: Use the Kb for ammonia to find the pOH, then convert to pH. (b) to (c) Buffer region: Calculate the pH using the Henderson-Hasselbalch equation, where pH = pKa + log{[A-]/[HA]} (d) to (f) Excess HCl region: Calculate the concentration of excess H+ and find the pH directly. Here's a detailed solution for each given point in the titration: (a) 0 mL HCl: Kb for NH3 = \(1.8 \times 10^{-5}\) \(K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\) We can use an ICE table to find the equilibrium concentrations: Initial: [NH3] = 0.030 M, [OH-] = 0, [NH4+] = 0 Change: -x, +x, +x Equilibrium: (0.030-x), x, x Plugging equilibrium concentrations into the Kb expression: \(1.8 \times 10^{-5} = \frac{x^2}{0.030-x}\) Solve for x (assuming x<<0.030): x ≈ 0.00219 = [OH-] pOH = -log(0.00219) ≈ 2.66 pH = 14 - pOH = 11.34 (b) 10.0 mL HCl: moles of HCl added = 0.00025 mol moles of NH3 initially = (0.030 mol/L) × (30.0 mL) × (1 L/1000 mL) = 0.00090 mol After reaction: moles of NH3 remaining = 0.00090 - 0.00025 = 0.00065 mol moles of NH4+ produced = 0.00025 mol Now, the total volume of the solution is 30 mL (NH3) + 10 mL (HCl) = 40 mL \[ NH_3 \] concentration = (# of moles of NH3) / (0.040 L) = 0.00065 mol/0.040 L = 0.01625 M \[ NH_4^+ \] concentration = (# of moles of NH4+) / (0.040 L) = 0.00025 mol/0.040 L = 0.00625 M Using Henderson-Hasselbalch equation (pKa of NH4+ is 9.25): pH = 9.25 + log(\(\frac{[NH_3]}{[NH_4^+]}\)) = 9.25 + log(\(\frac{0.01625}{0.00625}\)) ≈ 10.07 (c) 20.0 mL HCl: Follow the same process as in (b) -> pH ≈ 9.25 (d) 35.0 mL HCl: Follow the same process as in (b) -> At the equivalence point, moles of NH3 = moles of HCl. Calculate the concentration of NH4+ and use its Ka to find the pH. -> pH ≈ 5.96 (e) 36.0 mL HCl: Calculate the excess moles of HCl, determine the concentration of excess H+, and calculate pH directly -> pH ≈ 5.65 (f) 37.0 mL HCl: Follow the same process as in (e) -> pH ≈ 5.35 Final pH values for each point in the titration: (a) 11.34 (b) 10.07 (c) 9.25 (d) 5.96 (e) 5.65 (f) 5.35

Key concepts

pH Calculation

Understanding how to calculate the pH during a titration is crucial for grasping the changes that occur in a chemical reaction. pH is a measure of the acidity or basicity of a solution. It is calculated using the formula:\[pH = -\log[H^+]\]In titrations, especially those involving weak bases like ammonia (\(\text{NH}_3\)), the pH varies as more acid is added. You begin by calculating the pH based on the initial concentration of the solution, which often involves the dissociation constant of the weak base (\(K_b\)). As the titration progresses, you'll use different approaches:- Initially, calculate pH using the base dissociation.- In the buffer region, use the Henderson-Hasselbalch equation.- After reaching the equivalence point, focus on the acid formed and its dissociation.Each method reflects the unique characteristics of the solution's current state.

Buffer Solution

A buffer solution is a mixture that can resist changes in pH when small amounts of an acid or a base are added to it. In our titration example, ammonia (\(\text{NH}_3\)) and its conjugate acid, ammonium (\(\text{NH}_4^+\)), form a buffer solution. This occurs in the middle stages of the titration when neither the base nor the acid is in large excess.- Buffer solutions contain both the weak base and its conjugate acid at significant concentrations.- They maintain the pH of a solution within a narrow range, which is particularly useful during titration as analyte and titrant quantities adjust.Buffers are essential in biological systems and industrial applications where a stable pH is necessary. In titration, identifying the buffer region is important for accurate pH measurement and analysis.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a handy tool for estimating the pH of buffer solutions. It is expressed as:\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]Where:- \(pK_a\) is the negative log of the acid dissociation constant.- \([A^-]\) and \([HA]\) are the concentrations of the base and acid forms, respectively.During a titration, this equation simplifies the calculation of pH in the buffer region. Using the initial concentrations and the changes from the added titrant, you substitute into the equation to determine the current pH:- Calculate the molarity of the base (\(\text{NH}_3\)) and its conjugate acid (\(\text{NH}_4^+\)).- Use the known \(pK_a\) of the conjugate acid (ammonium in this case) to find pH.This equation supports a better understanding of pH changes in titration beyond simple stoichiometric calculations.

Equivalence Point

The equivalence point in a titration is where the amount of titrant, typically an acid, added is just enough to completely neutralize the base in the solution. This is a critical moment where the initial acid or base present has reacted fully: - **In strong acid-weak base titrations:** The equivalence point is reached when all the base is converted to its conjugate acid. - **At this point:** The pH does not equal 7, as would be expected in a strong acid-strong base titration, because the solution contains a weak acid. To find the equivalence point during a weak base titration, focus on calculating the number of moles of titrant added until they equal the moles of the original base. After the equivalence point, additional titrant contributes excess hydrogen ions, lowering the pH further. Understanding this point is crucial for reaching the desired acidity in industrial reactions or precise calculations in laboratory settings.