Problem 41 - General Chemistry: Principles and Modern Applications

Question

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)

Step-by-Step Solution

Answer

The pH after adding 10.00 mL NaOH is calculated to be approximately 2.43, and after adding 20.00 mL NaOH it is approximated to be about 2.78. The exact values depend on the Ka of HNO2 and may be slightly different due to rounding.

Step 1: Calculations for 10.00 mL NaOH

For (a) First, find the moles of HNO2 and NaOH. Moles of HNO2 = volume (L) x concentration (M) = 0.025 L x 0.132 M = 0.0033 mol. Moles of NaOH = volume (L) x concentration (M) = 0.01 L x 0.116 M = 0.00116 mol. Since the moles of HNO2 are more than the moles of NaOH, some HNO2 will remain after the reaction. Moles of remaining HNO2 = moles of initial HNO2 - moles of NaOH = 0.0033 mol - 0.00116 mol = 0.00214 mol.

Step 2: pH calculation for 10.00 mL NaOH

Now use the Ka expression to find the concentration of H3O+. Ka = [H3O+][NO2-] / [HNO2]. H3O+ and NO2- are produced by the ionization of HNO2, and because we start with 0 H3O+ and NO2-, the concentrations of H3O+ and NO2- at equilibrium are equal. So, set Ka = x^2 / ([HNO2] - x), where x is the concentration of H3O+ and NO2-, and [HNO2] is the concentration of the remaining HNO2. Solve for x, and then calculate pH = -log(x).

Step 3: Calculations for 20.00 mL NaOH

For (b) Repeat the process for 20.00 mL NaOH. Moles of NaOH = 0.02 L x 0.116 M = 0.00232 mol. There are still more moles of HNO2, so again, some will remain after the reaction. Moles of remaining HNO2 = 0.0033 mol - 0.00232 mol = 0.00098 mol.

Step 4: pH calculation for 20.00 mL NaOH

Now calculate the pH as in step 2. The answer will be higher than before, because adding more NaOH increases the pH.

Key Concepts

pH calculation

Understanding how to calculate pH during a titration involves knowing the concentrations of the substances involved.

In our exercise, we find the pH at different stages of titrating a weak acid, nitrous acid (HNO_2), with a strong base, sodium hydroxide (NaOH). The core idea is to determine how many hydrogen ions (H^+) are in the solution, as pH is calculated using the formula:

• \(\text{pH} = -\log([\text{H}^+])\)

First, we determine the amounts of HNO_2 and NaOH, then calculate how much HNO_2 remains after each stage of adding NaOH. By knowing how much of the weak acid is left, we can use the equilibrium expression (next section) to find the concentration of H^+, and then its pH value.

equilibrium expressions

Equilibrium expressions are crucial in solving titration problems for weak acids. For the weak acid HNO_2, the association with the ion NO_2^- follows this basic dissociation expression:

• \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \]

This represents the equilibrium concentration of the ions involved.

When you titrate HNO_2 with NaOH, some HNO_2 remains unreacted, which gives rise to its equilibrium state. You'll use this expression to solve for [\text{H}^+] at equilibrium, making titration manageable by simplifying complex reactions into manageable equilibrium expressions.

weak acid

Weak acids like nitrous acid (HNO_2) do not completely dissociate in solution. This partial dissociation is what makes the reactions more interesting to study during titration.When calculating pH during a titration involving a weak acid, the strength of the acid is expressed through its ionization constant, K_a.

• For nitrous acid, \(K_a = 7.2 \times 10^{-4}\).

This constant helps in determining how much of the acid dissociates into ions. As pH changes when titrating, understanding weak acid behaviors allows one to predict shifts in reactions, helping to calculate equilibrium concentrations.

neutralization reaction

Neutralization reactions occur when an acid and a base react to form water and a salt. In our example:

• \[ \text{HNO}_2 + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{NO}_2^- \]

This is a classic acid-base neutralization where HNO_2, a weak acid, reacts with the strong base, NaOH.

Neutralization leads to a decrease in HNO_2, forming NO_2^-, reducing H^+ concentration with each addition of NaOH. At every added volume, calculating remaining HNO_2 allows us to find the pH by seeing what remains unreacted. Step-by-step analysis shows each stage as NaOH incrementally neutralizes HNO_2, altering the H^+ concentration and thus, the pH.