Question

Fifty \(\mathrm{cm}^{3}\) of \(1.000 \mathrm{M}\) nitrous acid is titrated with \(0.850 \mathrm{M} \mathrm{NaOH}\). What is the \(\mathrm{pH}\) of the solution (a) before any \(\mathrm{NaOH}\) is added? (b) at half-neutralization? (c) at the equivalence point? (d) when \(0.10 \mathrm{~mL}\) less than the volume of \(\mathrm{NaOH}\) to reach the equivalence point is added? (e) when \(0.10 \mathrm{~mL}\) more than the volume of \(\mathrm{NaOH}\) to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure \(14.10(\mathrm{pH}\) versus volume \(\mathrm{NaOH}\) added).

Short answer

Question: Calculate the pH of a 1.00 M nitrous acid (HNO₂) solution at various stages during titration with sodium hydroxide (NaOH). (a) Initial pH (b) pH at half-neutralization (c) pH at the equivalence point (d) pH when 0.10 mL less than the volume of NaOH to reach the equivalence point is added (e) pH when 0.10 mL more than the volume of NaOH to reach the equivalence point is added (f) Construct a pH vs volume of NaOH added graph

Step-by-step solution

(a) Calculate the initial pH

First, we need to determine the initial pH of the 1.000 M nitrous acid solution before any NaOH is added. The dissociation of nitrous acid is given by the equation: HNO₂ ⇌ H⁺ + NO₂⁻. We can use the acid dissociation constant (Kₐ) of nitrous acid, which is 4.5 × 10⁻⁴. We can set up an ICE (Initial, Change, Equilibrium) table with the concentrations of the species and use the Kₐ expression to solve for the equilibrium concentration of H⁺ ions. Initial: [HNO₂] = 1.000 M, [H⁺] = 0, [NO₂⁻] = 0 Change: -x, +x, +x Equilibrium: (1.000 - x), x, x Now, we can set up the Kₐ expression: Kₐ = (x)(x) / (1.000 - x) 4.5 × 10⁻⁴ = x² / (1.000 - x) The concentration x is very small compared to 1.000, so we can approximate: 4.5 × 10⁻⁴ = x² / 1.000 After solving, x = [H⁺] = 0.0212 M Finally, we can calculate the initial pH: pH = -log([H⁺]) = -log(0.0212) ≈ 1.67

(b) Calculate the pH at half-neutralization

At half-neutralization, we have reacted half of the initial amount of nitrous acid with sodium hydroxide. The concentration of nitrous acid and its conjugate base (NO₂⁻) are equal at this point ([HNO₂] = [NO₂⁻]). Using the Kₐ expression again: Kₐ = ([H⁺][NO₂⁻]) / [HNO₂] 4.5 × 10⁻⁴ = ([H⁺][NO₂⁻]) / [NO₂⁻] [H⁺] = 4.5 × 10⁻⁴ M The pH at half-neutralization is: pH = -log(4.5 × 10⁻⁴) ≈ 3.35

(c) Calculate the pH at the equivalence point

At the equivalence point, all the nitrous acid has reacted with sodium hydroxide, and the only species left in the solution are water and the nitrite ion (NO₂⁻). The NO₂⁻ ion can undergo hydrolysis with water, leading to the formation of OH⁻ ions: NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻ We can use the base dissociation constant (Kₐ) for this reaction, which can be determined from the Kₐ of the nitrous acid: K₆ = Kₕ / Kₐ = 1 × 10⁻¹⁴ / 4.5 × 10⁻⁴ = 2.22 × 10⁻¹¹ Now, we can set up another ICE table for this equilibrium and solve for the equilibrium concentration of OH⁻ ions. Initial: [OH⁻] = 0, [HNO₂] = 0, [NO₂⁻] = 50.0 mmol / (50.0 + Veq mL) Change: +x, +x, -x Equilibrium: x, x, (50.0 mmol / (50.0 + Veq mL) - x) K₆ = (x)(x) / ((50.0 mmol / (50.0 + Veq mL) - x)) = 2.22 × 10⁻¹¹ x = [OH⁻] = 0.00000297 M To find the pH at the equivalence point, we first find the pOH: pOH = -log([OH⁻]) = 5.53 Then we find the pH: pH = 14 - pOH = 14 - 5.53 = 8.47

(d) Calculate the pH when 0.10 mL less than the volume of NaOH to reach the equivalence point is added

First, we need to find the remaining amount of nitrous acid in mol (HNO₂) by subtracting the number of moles of NaOH added: nHNO2 = 1.0 M × 50.0 mL - 0.85 M × (Veq - 0.10 mL). Then, we can again apply the Kₐ expression for nitrous acid, using concentrations of all three species in equilibrium, and solve for [H⁺] ions concentration. Finally, calculate the pH using the found [H⁺] concentration.

(e) Calculate the pH when 0.10 mL more than the volume of NaOH to reach the equivalence point is added

First, we need to find the additional amount of NaOH in mol (nOH⁻) added: nOH⁻ = 0.85 M × (Veq + 0.10 mL) - (1.0 M × 50.0 mL). Then, the concentration of OH⁻ ions in equilibrium is approximated to be equal to nOH⁻ / (50.0 mL + Veq + 0.10 mL). Calculate pOH using -log([OH⁻]), and find the pH using the relationship pH = 14 - pOH.

(f) Construct a pH vs volume of NaOH added graph

Using the calculated pH values, plot a graph similar to Figure 14.10 with volume of NaOH added (in mL) along the x-axis and pH along the y-axis. The graph will show an initial pH of 1.67 before any NaOH is added, a rapid rise in pH as NaOH is added until it reaches a pH of 3.35 at half-neutralization, followed by a more gradual increase until reaching a pH of 8.47 at the equivalence point, and then another rapid rise in pH as more NaOH is added past the equivalence point.

Key concepts

Acid-Base Titration

Understanding acid-base titration is crucial for students who study chemistry. In simple terms, it's a technique used to find the concentration of a substance by reacting it with another substance of known concentration, called the titrant. Typically, one of the substances is an acid and the other a base.

During the titration process, small volumes of the titrant are added to the substance being measured, usually with a burette. As the acid and base react, they form water and a salt in a neutralization reaction. The moment when all the acidic or basic sites have reacted is called the equivalence point, which is key in titration calculations. A pH indicator or a pH meter can be used to tell when this point has been reached. By calculating the volume of titrant used to reach the equivalence point, the concentration of the unknown substance can be determined. This concept helped students to follow the step by step titration exercise by revealing the underlying science.

pH Calculation

The pH scale measures how acidic or basic a solution is, ranging from 0 (very acidic) to 14 (very alkaline), with 7 being neutral. pH is calculated using the formula pH = -log([H⁺]), where [H⁺] is the concentration of hydrogen ions in moles per liter.

In the exercise, students used this formula to calculate the pH of the nitrous acid solution before titration started, at half-neutralization, and at various points during the titration process. The pH calculation is vital for determining how the acidity changes as the base (NaOH) is added. This is used, for example, to find out the initial pH of the acid solution or the pH at the half-neutralization where the concentrations of the acid and its conjugate base are equal.

Equivalence Point

The equivalence point is a fundamental concept in titrations, particularly those involving acids and bases. It refers to the point during the titration where the number of moles of hydrogen ions (H⁺) equals the number of moles of hydroxide ions (OH⁻). At this juncture, the chemical reaction between the acid and base is stoichiometrically complete.

Identification of Equivalence Point

In the exercise, students had to calculate the volume of NaOH needed to reach the equivalence point. This is critical for analyzing titration curves and understanding the chemistry of acids and bases. The equivalence point can be identified using indicators, which change color at a certain pH range, or by using a pH meter, which provides a more precise measurement of the pH changes that occur during titration.

The pH values at different stages, including the equivalence point, provide insight into the properties of the acid and base. For example, in a weak acid-strong base titration, the equivalence point pH will be above 7, indicating the formation of a basic solution due to the presence of the conjugate base of the weak acid. The step-by-step solution in the exercise illustrates the importance of calculating the equivalence point accurately to understand the reaction's completion.

Acid Dissociation Constant

The acid dissociation constant, K_a, is a value that measures the strength of an acid in solution. It's the equilibrium constant for the dissociation of the acid into its conjugate base and a hydrogen ion. The larger the K_a, the stronger the acid, and the more it will dissociate.

In the given problem, students work with nitrous acid (HNO₂), which dissociates to form H⁺ and NO₂^-. The K_a for nitrous acid is a key element in determining the pH of the solution at various points during the titration. By applying the formula K_a = ([H⁺][NO₂⁻])/[HNO₂], students can calculate the amount of hydrogen ions present in the solution and thus find the pH. An understanding of the K_a concept also helps when considering the reverse reaction at the equivalence point, where the formation of hydroxide ions from the conjugate base (NO₂⁻) must be examined.