Question

Repeat the procedure in Exercise 61 , but for the titration of \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with \(0.100 \mathrm{M}\) \(\mathrm{HCl}\)

Short answer

The pH during the titration of 25.0 mL 0.100 M NH3 with 0.100 M HCl at various stages is as follows: \(1.\) Before titration, pH = 11.13, \(2.\) After adding 5.0 mL HCl, pH ≈ 11.14, \(3.\) At the equivalence point, pH = 4.82, and \(4.\) After adding 35.0 mL HCl, pH = 1.78.

Step-by-step solution

Calculate initial pH (before titration begins)

Determine the initial concentration of NH3 and use the Kb to calculate the equilibrium concentrations of NH3, NH4+ and OH-, and from there, determine the initial pH using the pOH. Initial concentration of NH3 = 0.100 M Kb = 1.8 x 10^(-5) Set up an ICE table for the reaction: NH3 + H2O <-> NH4+ + OH- I 0.100 0 0 C -x x x E 0.100-x x x Kb = [NH4+][OH-] / [NH3] 1.8 x 10^(-5) = x^2 / (0.100-x) Approximate that x << 0.100, so 0.100-x ≈ 0.100: x ≈ √(1.8 x 10^(-5) * 0.100) = 1.34 x 10^(-3) M (concentration of OH-) Now, calculate pOH and pH: pOH = -log(1.34 x 10^(-3)) = 2.87 pH = 14 - pOH = 14 - 2.87 = 11.13

Calculate pH after adding a small amount of titrant

Add 5.0 mL of 0.100 M HCl to the 25.0 mL of 0.100 M NH3. Moles of NH3 initially present = 0.100 M * 0.025 L = 0.00250 mol Moles of HCl added = 0.100 M * 0.005 L = 0.00050 mol A stoichiometry calculation shows that 0.0005 moles of NH3 will be left after the reaction with HCl: NH3 remaining after reaction with HCl = 0.00250 - 0.00050 = 0.00200 mol New concentration of NH3: 0.00200 / 0.030 L = 0.0667 M (after dilution with 5.0 mL HCl) Now, use the Kb equation to determine the new [OH-], pOH, and pH.

Calculate pH at the equivalence point

At the equivalence point, all NH3 has reacted with HCl, so we have an NH4Cl solution. The NH4+ concentration is: [ NH4+] = moles of NH4+ / total volume = 0.00250 / 0.030 L = 0.0833 M Use the Kb for NH3 to find the Ka for NH4+: Ka = Kw / Kb = 1 x 10^(-14) / (1.8 x 10^(-5)) = 5.56 x 10^(-10) Now, calculate the hydrogen ion concentration, [H+], using the Ka as follows: Ka = [H+][Cl-]/[NH4+] [H+]^2 = Ka * [NH4+] Approximate [H+] ≈ 1.53 x 10^(-5) M Now, calculate the pH: pH = -log(1.53 x 10^(-5)) = 4.82

Calculate pH after the equivalence point

Adding more HCl after the equivalence point will result in the solution becoming acidic. Let's calculate the pH after adding 35.0 mL of 0.100 M HCl. Moles of HCl added = 0.100 M * 0.035 L = 0.00350 mol Excess moles of HCl = 0.00350 - 0.00250 = 0.00100 mol Total volume after adding HCl = 25.00 mL + 35.00 mL = 60.00 mL = 0.060 L Excess HCl concentration = 0.00100 mol / 0.060 L = 0.0167 M pH = -log(0.0167) = 1.78

Key concepts

Titration of NH3 with HCl

Titration is a laboratory process where a solution known as a titrant is slowly added to another solution with an unknown concentration until a reaction completion point is reached, often indicated by a color change. In the case of titrating ammonia (NH3) with hydrochloric acid (HCl), we have a classic acid-base neutralization.

Ammonia is a weak base and reacts with the strong acid, HCl, to produce the ammonium ion, NH4+, and chloride, Cl-. This titration proceeds in a stepwise fashion:

• Before any HCl is added, the NH3 solution has a basic pH due to the presence of OH- ions that form when NH3 reacts with water.
• Upon adding HCl, NH3 gets neutralized, forming NH4+, decreasing the solution's pH gradually.
• At the equivalence point, all NH3 has been converted to NH4+, marking the point of complete neutralization.
• Any additional HCl after the equivalence point increases the acidity of the solution, resulting in a lower pH.

Understanding these phases is crucial for analyzing the titration curve and determining various solution characteristics during the process.

Calculation of pH in Titration

Calculating the pH at various stages of a titration involves understanding the interactions between the acid and base and the resulting changes in ion concentration.

Initially, we find the pH of the ammonia solution before adding any HCl by using the equilibrium constant (Kb) for NH3 to find the hydroxide ion (OH-) concentration. Once HCl is added, the balance shifts, and pH calculations must account for the NH3 neutralized by the HCl. As more HCl is added, up until the equivalence point, fewer NH3 molecules are available to form OH-, hence the pH decreases.

At the equivalence point, the pH is determined by the ammonium ion's ability to hydrolyze to form NH3 and H+. Beyond the equivalence point, the pH of the solution is dictated by the concentration of excess HCl, turning the solution acidic.

Equivalence Point

The equivalence point in a titration is the moment when the number of moles of the titrant added equals the number of moles of substance present in the solution being titrated—in this case, when the moles of HCl equal the moles of NH3.

For ammonia titration with hydrochloric acid, the equivalence point typically occurs at a pH less than 7 because NH4+ (the conjugate acid of NH3) is a weak acid and can slightly dissociate to form H+ ions, which lowers the pH of the solution.

Finding the equivalence point is vital as it provides valuable information needed to determine the original concentration of the substance being titrated, in this case, NH3, and is often determined using a pH meter or an indicator that changes color at a specific pH range.

Buffer Solution pH Calculation

A buffer solution is one that can resist changes in pH upon the addition of small amounts of acid or base. During the titration of NH3 with HCl, before reaching the equivalence point, the mixture acts as a buffer because it contains significant amounts of both NH3 and its conjugate acid, NH4+.

The pH of a buffer can be calculated using the Henderson-Hasselbalch equation for basic buffers: \[ \text{pH} = \text{pKb} + \log\left(\frac{\text{[base]}}{\text{[conjugate acid]}}\right) \] where [base] is the concentration of NH3 and [conjugate acid] is the concentration of NH4+. By calculating the ratio of these two species at any point during the titration prior to the equivalence point, one can determine the pH of the buffer solution. This calculation is helpful when working with systems that need to maintain a relatively constant pH, such as biological systems or certain industrial processes.