Question

A \(50.0 \mathrm{~mL}\) sample of \(0.250 \mathrm{M}\) ammonia \(\left(\mathrm{NH}_{3}, K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) is titrated with \(0.250 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) after the addition of each of the following volumes of acid: (a) \(0.0 \mathrm{~mL}\) (b) \(25.0 \mathrm{~mL}\) (c) \(50.0 \mathrm{~mL}\) (d) \(60.0 \mathrm{~mL}\)

Short answer

(a) pH = 11.32, (b) pH = 9.26, (c) pH = 5.13, (d) pH = 1.64.

Step-by-step solution

Initial Information

We begin with a solution containing \(50.0\, \text{mL}\) of \(0.250\, \text{M}\) ammonia (\(\text{NH}_3\)) with a base ionization constant \(K_b = 1.8 \times 10^{-5}\). We will titrate this with \(0.250\, \text{M}\) nitric acid (\(\text{HNO}_3\)).

Calculate Initial pH (0.0 mL Add Acid)

For ammonia, calculate the hydroxide ion concentration using the base dissociation.\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]Set up an ICE table, solve for \([\text{OH}^-]\), then calculate \(\text{pOH}\) and finally \(\text{pH}\). \[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \approx x^2 / (0.250 - x) \] Since \(x\) is small, \(0.250 - x \approx 0.250\): \[ x = \sqrt{K_b \cdot 0.250} = \sqrt{1.8 \times 10^{-5} \cdot 0.250} = 0.0021 \] \[ \text{pOH} = -\log(0.0021) \approx 2.68 \] \[ \text{pH} = 14 - \text{pOH} = 11.32 \]

Determine pH After Adding 25.0 mL Acid

This is the half-equivalence point where half of the base has reacted. Thus, \([\text{NH}_3] = [\text{NH}_4^+]\), and \(\text{pH} = \text{pK}_b\).\[ \text{pK}_b = -\log(1.8 \times 10^{-5}) \approx 4.74 \]Calculate \(\text{pK}_w - \text{pK}_b\) for \(\text{pK}_a\), where \(\text{pK}_a = 9.26 \), and thus, \(\text{pH} = 9.26\).

Calculate pH At Equivalence Point (50.0 mL Acid)

At this point, all \(\text{NH}_3\) is converted to \(\text{NH}_4^+\). Only the \(\text{NH}_4^+\) ion affects the \(\text{pH}\).Calculate using \(K_a\) of \(\text{NH}_4^+\):\[ K_w = 1.0 \times 10^{-14}, K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]\[ \text{pH} = 7 - 0.5 \cdot \log(5.56 \times 10^{-10}) \approx 5.13 \]

Calculate pH Beyond Equivalence Point (60.0 mL Acid)

Here, excess \(\text{HNO}_3\) will be present. Calculate \([\text{H}^+]\) from excess acid.Find mmol of acid:\[ 0.250 \text{M} \times (60.0 - 50.0) \mathrm{~mL} = 2.5 \text{ mmol} \]Final total volume = \(110.0 \text{ mL}\), so\[ [\text{H}^+] = \frac{2.5 \times 10^{-3} \text{ mol}}{0.110 \text{ L}} = 0.0227 \text{ M} \]\(\text{pH} = -\log(0.0227) = 1.64\).

Key concepts

pH calculation

Calculating the pH in a solution involves understanding the concentration of hydrogen ions (\([ ext{H}^+]\)) in the solution. For a basic solution like ammonia (\( ext{NH}_3\)), we initially calculate the pH by determining the hydroxide ion concentration (\([ ext{OH}^-]\)) because ammonia is a weak base that produces \( ext{OH}^-\) ions when dissolved in water.

Start by setting up an ICE (Initial, Change, Equilibrium) table to express the changes in concentrations during the reaction. Once you solve for \([ ext{OH}^-]\), convert this to pOH by using \( ext{pOH} = - ext{log}([ ext{OH}^-])\).

Finally, because pH and pOH are related by the equation \( ext{pH} + ext{pOH} = 14\), you can determine the pH using: \( ext{pH} = 14 - ext{pOH}\). Calculating the pH at different stages of titration helps you understand how the concentration of \([ ext{OH}^-]\) and \([ ext{H}^+]\) change as acid is added.

equivalence point

The equivalence point in an acid-base titration is when the number of moles of acid equals the number of moles of base. In this specific problem, when 50.0 mL of nitric acid (\( ext{HNO}_3\)) is added to 50.0 mL of ammonia (\( ext{NH}_3\)), the moles of acid equal the moles of base.

At the equivalence point, all the \( ext{NH}_3\) has reacted to form \( ext{NH}_4^+\) ions. Therefore, the pH is determined by the acidic nature of the ammonium ion (\( ext{NH}_4^+\)), which raises the concentration of \([ ext{H}^+]\) ions and shifts the pH below 7.

This is why at this stage, we need to calculate the pH using the \(K_a\) of \( ext{NH}_4^+\), representing the weak conjugate acid formed from the neutralization reaction.

base ionization constant

The base ionization constant, \(K_b\), is crucial in understanding how well a base dissociates into ions in water. For ammonia, \(K_b = 1.8 \times 10^{-5}\) indicates that it weakly dissociates to form \( ext{NH}_4^+\) and \( ext{OH}^-\) ions.

Using \(K_b\), we can set up the dissociation equation, calculate the concentration of \([ ext{OH}^-]\), and determine the \( ext{pOH}\). This, in turn, helps calculate the initial pH of the ammonia solution before any titration begins.

The \(K_b\) value also plays a role in calculating the \(K_a\) of the conjugate acid (\( ext{NH}_4^+\)) using the relation \(K_w = K_a \times K_b\), where \(K_w\) is the ion-product constant of water (\(1.0 \times 10^{-14}\) at 25°C).

neutralization reaction

A neutralization reaction occurs during titration when an acid reacts with a base to form water and a salt. In our ammonia titration problem, \( ext{HNO}_3\) (an acid) reacts with \( ext{NH}_3\) (a base) to form \( ext{NH}_4^+\) and \( ext{NO}_3^-\): \( ext{NH}_3 + ext{HNO}_3 \rightarrow ext{NH}_4^+ + ext{NO}_3^-\).

During this reaction, \( ext{NH}_4^+\) ions are produced, which affect the pH of the solution as they form a weak acid. The stage at which all the \( ext{NH}_3\) has been converted into \( ext{NH}_4^+\) marks the equivalence point.

Once the acid added surpasses the initial base amount, the solution becomes more acidic, hence the pH decreases. Understanding this process helps to predict and calculate how the pH changes throughout the titration.