Problem 29 - Chemistry and Chemical Reactivity

Question

You titrate \(25.0 \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{NH}_{3}\) with \(0.10 \mathrm{M} \mathrm{HCl}\). (a) What is the pH of the NH \(_{3}\) solution before the titration begins? (b) What is the \(\mathrm{pH}\) at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) What indicator in Figure 18.10 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00 \(15.0,20.0,22.0,\) and \(30.0 \mathrm{mL}\) of the acid. Combine this information with that in parts (a)-(c) and plot the titration curve.

Step-by-Step Solution

Answer

a) 11.13; b) 5.28; c) 9.25; d) Bromocresol green; e) titration curve varies with volumes.

Step 1: Calculate Initial pH of NH_3 Solution (Part a)

To find the initial pH of the NH\(_3\) solution, we use the formula for weak base dissociation. The equation for the dissociation of ammonia is \( \mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^- \). The base dissociation constant \( K_b \) for ammonia is \( 1.8 \times 10^{-5} \).Use the formula:\[ \mathrm{OH}^- = \sqrt{K_b \times [\mathrm{NH}_3]} \]\[ \mathrm{OH}^- = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \]Calculate \( \mathrm{pOH} \):\[ \mathrm{pOH} = -\log{(1.34 \times 10^{-3})} \approx 2.87 \]Then calculate \( \mathrm{pH} \) using \( \mathrm{pH} + \mathrm{pOH} = 14 \):\[ \mathrm{pH} = 14 - 2.87 = 11.13 \]

Step 2: Calculate pH at Equivalence Point (Part b)

At the equivalence point, all \( \mathrm{NH}_3 \) has been converted to \( \mathrm{NH}_4^+ \), a weak acid. Calculate the concentration of \( \mathrm{NH}_4^+ \): Solution volume at equivalence is \( 25.0 \mathrm{mL} + 25.0 \mathrm{mL} = 50.0 \mathrm{mL} \).Moles of \( \mathrm{NH}_4^+ \):\[ 0.0025 \mathrm{moles} \quad \text{(since initial volume was } 25.0 \mathrm{mL})\]Concentration \( [\mathrm{NH}_4^+] = \frac{0.0025}{0.050} = 0.05 \mathrm{M} \).Using \( K_a \):\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]Use the formula for weak acid dissociation:\[ [\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{NH}_4^+]} = \sqrt{5.56 \times 10^{-10} \times 0.05} \approx 5.27 \times 10^{-6} \]Calculate \( \mathrm{pH} \):\[ \mathrm{pH} = -\log{(5.27 \times 10^{-6})} \approx 5.28 \]

Step 3: Calculate pH at Halfway Point (Part c)

At the halfway point, half the \( \mathrm{NH}_3 \) is converted to \( \mathrm{NH}_4^+ \), and \( [\mathrm{NH}_3] = [\mathrm{NH}_4^+] \).Using the Henderson-Hasselbalch equation:\[ \mathrm{pH} = pK_a + \log{\left(\frac{[\mathrm{base}]}{[\mathrm{acid}]}\right)} \]At halfway point, \( \[ \mathrm{pH} = pK_a = 9.25 \] \) because \( [\mathrm{base}] = [\mathrm{acid}] \).

Step 4: Select Proper Indicator (Part d)

Choose an indicator with a transition range that includes the pH at the equivalence point, around 5.28. Methyl orange which has a pH range of 3.1 to 4.4 or bromocresol green with a pH range of 4.0 to 5.6 would likely show the color change around the equivalence point to facilitate detection.

Step 5: Calculate pH after Adding Acid to Titrate (Part e)

Determine the moles of \( \mathrm{NH}_3 \), \( \mathrm{HCl} \), and the resulting pH after adding different volumes.1. **5.0 mL \( \mathrm{HCl} \):**Moles \( \mathrm{HCl} = 0.5 \times 10^{-3} \); \( [\mathrm{NH}_3] = 0.0025 - 0.0005 = 0.002 \) moles remain.Partial neutralization; use Henderson-Hasselbalch:\[ \mathrm{pH} = 9.25 + \log{\left(\frac{0.002}{0.0005}\right)} \approx 10.06 \]2. **15.0 mL \( \mathrm{HCl} \):**Moles \( \mathrm{HCl} = 0.0015 \); \( [\mathrm{NH}_3] = 0.0025 - 0.0015 = 0.001 \) moles remain.Use Henderson-Hasselbalch:\[ \mathrm{pH} = 9.25 + \log{\left(\frac{0.001}{0.0015}\right)} \approx 9.05 \]3. **20.0 mL \( \mathrm{HCl} \):**Moles \( \mathrm{HCl} = 0.002 \); equal amounts exist.Using Henderson-Hasselbalch:\[ \mathrm{pH} = 9.25 \quad \text{(at halfway)} \]4. **22.0 mL \( \mathrm{HCl} \):**Moles \( \mathrm{HCl} = 0.0022 \); \( [\mathrm{NH}_4^+] = 0.0002 \), suppresses \( \mathrm{H}^+ \), find \([\mathrm{OH}^-]\) again. \( \mathrm{pH} \approx 8.5 \)5. **30.0 mL \( \mathrm{HCl} \):**Beyond equivalence point, find excess \( \mathrm{H}^+ \):\[ \mathrm{pH} = -\log{\left(\frac{0.0005}{0.055}\right)} \approx 2.65 \]Plot these values accordingly.

Key Concepts

pH calculation

Calculating pH is crucial in understanding acid-base reactions. pH measures how acidic or basic a solution is. For a weak base like ammonia (\(\text{NH}_3\)), the initial pH is determined by its dissociation in water. The base accepts a proton to form \(\text{NH}_4^+\) and \(\text{OH}^-\). We use the base dissociation constant \(K_b\) to find \([\text{OH}^-]\) and calculate \(\text{pOH}\):

• Use \([\text{OH}^-] = \sqrt{K_b \times [\text{NH}_3]}\).
• Calculate \(\text{pOH} = -\log{[\text{OH}^-]}\).
• Find \(\text{pH} = 14 - \text{pOH}\).

This helps us determine the pH before the titration begins, giving insight into the solution's initial acidic or basic nature.

weak base

A weak base partially ionizes in water, unlike strong bases. Ammonia (\(\text{NH}_3\)) serves as a classic example. When dissolved, it forms a dynamic equilibrium where only some molecules become \(\text{NH}_4^+\) and \(\text{OH}^-\).
Key characteristics include:

• Low \(K_b\) value, indicating limited ionization.
• Aqueous solutions with higher pH than neutral water but not as high as a strong base.
• Reacts with added acids like \(\text{HCl}\) to form a conjugate acid (\(\text{NH}_4^+\)), affecting the pH during titration.

Understanding these properties enables prediction of pH changes during titration, especially at different points like halfway or equivalence.

equivalence point

The equivalence point in titration is where the amount of titrant added fully reacts with the analyte. For ammonia titrated with \(\text{HCl}\), the equivalence occurs when all \(\text{NH}_3\) is converted to \(\text{NH}_4^+\).
Steps to approach this concept:

• Recognize the solution is now acidic due to the formation of \(\text{NH}_4^+\), a weak acid.
• Calculate concentration of \(\text{NH}_4^+\) using moles and total volume.
• Use \(K_a\) derived from \(K_w/K_b\) to find \([\text{H}^+]\) and determine pH.

This point is crucial as the pH indicates the completion of reaction, which guides the selection of suitable indicators.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a helpful tool in calculating pH, especially for buffer solutions. It's essential during the titration of weak bases. It looks like:
\[\text{pH} = pK_a + \log{\left(\frac{[\text{base}]}{[\text{acid}]}\right)}\]At the halfway point in titration:

• Half of \(\text{NH}_3\) converts to \(\text{NH}_4^+\), making \([\text{base}] = [\text{acid}]\).
• The equation simplifies to \(\text{pH} = pK_a\).
• This equality simplifies calculations and helps understand solution stability.

The equation highlights the balance between the base and its conjugate acid, allowing precise control over pH in many scenarios.