Question

A \(35.00-\mathrm{mL}\) sample of \(0.487 \mathrm{M} \mathrm{KBrO}\) is titrated with \(0.264 \mathrm{M} \mathrm{HNO}_{3} \cdot\left(K_{\mathrm{b}} \mathrm{BrO}^{-}=4.0 \times 10^{-6}\right)\) (a) Write a balanced net ionic equation for the reaction. (b) How many milliliters of \(\mathrm{HCl}\) are required to reach the equivalence point? (c) What is the \(\mathrm{pH}\) at the equivalence point? (d) Calculate \(\left[\mathrm{K}^{+}\right],\left[\mathrm{NO}_{3}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{BrO}^{-}\right],\) and \([\mathrm{HBrO}]\) at the equivalence point. (Assume volumes are additive.)

Short answer

Answer: At the equivalence point, the ion concentrations are as follows: [K+] = 0.487 mol/L [NO3-] = 0.264 mol/L [H+] = 3.246 × 10^(-5) mol/L [BrO-] = 0 mol/L [HBrO] = 0.215 mol/L

Step-by-step solution

1. Write the balanced net ionic equation.

To write the balanced net ionic equation, we first identify the reactants and products. In this case, the weak base KBrO reacts with the strong acid HNO\(_3\). KBrO(aq) + HNO\(_3\)(aq) → KNO\(_3\)(aq) + HBrO(aq) Since KNO\(_3\) is a soluble salt, it will dissociate into ions in solution. HBrO is a weak acid and does not dissociate completely. So, the net ionic equation is: BrO\(^{-}\)(aq) + H\(^{+}\)(aq) → HBrO(aq)

2. Calculate the volume of HNO\(_3\) required to reach the equivalence point.

At the equivalence point, moles of acid are equal to moles of base. We can use the formula: M\(_1\)V\(_1\) = M\(_2\)V\(_2\) Where M\(_1\) and V\(_1\) are the concentration and volume of the weak base KBrO, and M\(_2\) and V\(_2\) are the concentration and volume of the strong acid HNO\(_3\). Plugging in the given values: (0.487 mol/L)(35.00 mL) = (0.264 mol/L)(V\(_2\)) Solving for V\(_2\): V\(_2\) = (0.487 mol/L × 35.00 mL) / (0.264 mol/L) = 64.57 mL So, 64.57 mL of HNO\(_3\) is required to reach the equivalence point.

3. Calculate the pH at the equivalence point.

At the equivalence point, all the BrO\(^-\) ions have reacted with the H\(^+\) ions to form HBrO, a weak acid. We can use the Ka expression of the weak acid to find the concentration of H\(^+\) ions at the equivalence point. Ka = (Kw/Kb) = ([H\(^+\)][BrO\(^-\)]) / [HBrO] To find Ka, we use the provided Kb value of BrO\(^-\) (4.0 × 10\(^{-6}\)) and the known value of Kw = 1.0 × 10\(^{-14}\). Ka = (1.0 × 10\(^{-14}\)) / (4.0 × 10\(^{-6}\)) = 2.5 × 10\(^{-9}\) Since the reaction produced equal moles of HBrO and KNO\(_3\), we can find their concentrations at the equivalence point: [HBrO] = (0.487 mol/L × 35.00 mL) / (35.00 mL + 64.57 mL) = 0.215 mol/L Now, solve for [H\(^+\)] using the Ka expression: 2.5 × 10\(^{-9}\) ≈ [(x)(x)] / (0.215 mol/L - x) Since x is likely very small compared to 0.215, we can approximate: 2.5 × 10\(^{-9}\) ≈ (x^2) / (0.215) x = √(2.5 × 10\(^{-9}\) × 0.215) = 3.246 × 10\(^{-5}\) Finally, we use the definition of pH to find the pH at the equivalence point: pH = -log([H\(^+\)]) = -log(3.246 × 10\(^{-5}\)) ≈ 4.49 The pH at the equivalence point is approximately 4.49.

4. Calculate ion concentrations at the equivalence point.

We can now find the concentrations of K\(^+\), NO\(_3^-\), H\(^+\), BrO\(^-\), and HBrO at the equivalence point. [K\(^+\)]: Since KBrO fully dissociated, [K\(^+\)] = [BrO\(^-\)], which reacted completely with HNO\(_3\) to form HBrO. Thus, the [K\(^+\)] will be equal to the initial concentration of BrO\(^-\): [K\(^+\)] = 0.487 mol/L [NO\(_3^-\)]: Since all the HNO\(_3\) was used to react with BrO\(^-\), the [NO\(_3^-\)] formed is equal to the initial [H\(^+\)]: [NO\(_3^-\)] = 0.264 mol/L [H\(^+\)]: We already calculated [H\(^+\)] to be 3.246 × 10\(^{-5}\) mol/L. [BrO\(^-\)]: At the equivalence point, no BrO\(^-\) is left, as it fully reacted with HNO\(_3\). [BrO\(^-\)] = 0 mol/L [HBrO]: As previously calculated, [HBrO] = 0.215 mol/L. At the equivalence point, the concentrations are: [K\(^+\)] = 0.487 mol/L [NO\(_3^-\)] = 0.264 mol/L [H\(^+\)] = 3.246 × 10\(^{-5}\) mol/L [BrO\(^-\)] = 0 mol/L [HBrO] = 0.215 mol/L

Key concepts

Net Ionic Equation

The net ionic equation is a simplified version of a chemical reaction that only shows the species that actually change during the reaction. In titrations involving a weak base and a strong acid, like the one with potassium bromate (KBrO) and nitric acid (HNO₃), the net ionic equation provides a clear picture of the acid-base neutralization that occurs.

For our given problem, the reaction between the weak base bromate ion (BrO^-) and the hydrogen ion (H⁺) from the strong acid forms hypobromous acid (HBrO). Here's the net ionic equation:
BrO^-(aq) + H⁺(aq) → HBrO(aq)

This equation leaves out the spectator ion, potassium (K⁺), which does not participate in the reaction. Understanding this helps you focus on the essence of the acid-base reaction and is particularly important when predicting the properties of the resulting solution, such as pH.

Equivalence Point

The equivalence point in a titration is the moment when the number of moles of acid equals the number of moles of base. At this juncture, the acid has completely neutralized the base, leading to a stoichiometrically balanced state. For our question involving KBrO titrated with HNO₃, the equivalence point is reached when the moles of BrO^- from KBrO react fully with the H⁺ from HNO₃.

Calculating the volume of acid needed to reach the equivalence point involves stoichiometric conversion using the initial concentrations and volumes. Remembering that concentration multiplied by volume gives the number of moles, you can use the equation:
M₁V₁ = M₂V₂

Here, M₁ and V₁ refer to the molarity and volume of the base (KBrO), while M₂ and V₂ refer to the molarity and volume of the acid (HNO₃). Once you solve for V₂, you have the volume of HNO₃ that results in stoichiometric neutrality - the essence of the equivalence point.

pH Calculation

The pH of a solution is a logarithmic measure of its hydrogen ion concentration and is a key concept in understanding the acidity or basicity of a solution. Calculating the pH at the equivalence point, particularly in a titration involving a weak base (like KBrO) and a strong acid (like HNO₃), can be challenging as the resulting solution will have a pH less than 7 due to the production of a weak acid (HBrO).

To calculate the pH at the equivalence point, we start by determining the molarity of the weak acid formed, then compute its dissociation into H⁺ ions using the Ka value, which can be derived from the provided Kb value and the water dissociation constant (Kw). Through the expression:
Ka ≈ ([H⁺]²) / [HBrO]

we estimate the concentration of hydrogen ions ([H⁺]) and then use the pH formula:
pH = -log([H⁺])

The calculation often involves approximations since the concentration of H⁺ ions is very small compared to the initial concentration of the weak acid. This understanding of the relationship between the weak acid's dissociation, hydrogen ion concentration, and pH is vital for analyzing the resulting solution's properties at the equivalence point.