Problem 107 - General Chemistry: Principles and Modern Applications

Question

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

Step-by-Step Solution

Answer

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 L of a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4}\mathrm{Cl}\) is to raise the pH very slightly.

Step 1: Determine reaction of buffer solution

When \(0.001 \mathrm{mol} \mathrm{KOH}\) (a strong base) is added to the buffer solution, it reacts with the weak acid (ammonium ion) in the buffer to form more ammonia and water. This is shown in the below equation \[ \mathrm{NH}_{4}^{+} + OH^{-} -> \mathrm{NH}_{3} + H_{2}O \]

Step 2: Calculate new concentrations of buffer components

Calculate the new concentrations of buffer components. The concentration of \( \mathrm{NH}_{4}^{+}\) ions would decrease by number of moles of \(KOH\) = \(0.10 - 0.001 = 0.099 M\). The concentration of \( \mathrm{NH}_{3}\) would increase by number of moles of \(KOH\) = \(0.10 + 0.001 = 0.101 M\).

Step 3: Use the Henderson-Hasselbalch equation to calculate change in pH

Now, use the Henderson-Hasselbalch equation for buffers, \(pH = pK_a + log10([base]/[acid])\). The \(pK_a\) value of \( \mathrm{NH}_{4}^{+}\) is 9.25. Substitute the new concentrations into the equation: \(pH = 9.25 + log10(0.101/0.099) = 9.25 + 0.00899 = 9.259. Thus, the pH slightly increases after addition of KOH.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a fundamental tool for anyone learning about buffer solutions in chemistry. It's used to estimate the pH of a buffer solution containing a weak acid and its conjugate base. Here's the equation: \[pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)\]Where:

• \(pH\) is the measure of acidity or basicity of the solution.

• \(pK_a\) is the negative log of the acid dissociation constant, a measure of the strength of the weak acid.

• \([\text{base}]\) is the concentration of the base form of the buffer.

• \([\text{acid}]\) is the concentration of the acid form of the buffer.

This equation is derived from the equilibrium expression for the dissociation of a weak acid, allowing us to relate pH changes to the ratio of the concentrations of the base and acid components in a buffer. It highlights how buffers resist pH changes by adjusting the acid and base ratio as acids or bases are added.

pH Calculation

Calculating pH in buffer solutions can seem tricky, but once you understand the process, it becomes straightforward. In a buffer solution, acids and bases are not completely dissociated, allowing for a stable pH. To calculate pH using the Henderson-Hasselbalch equation, you need the pKa of the weak acid and the concentrations of the acid and base molecules.

Take for instance ammonia (\(\mathrm{NH}_3\)), which forms a buffer with its conjugate acid, ammonium (\(\mathrm{NH}_4^+\)). If the concentration of \(\mathrm{NH}_3\) slightly increases and \(\mathrm{NH}_4^+\) slightly decreases due to added KOH, you can plug these values into the equation to find the new pH.

For example, given:

• The concentration of \(\mathrm{NH}_3\) is 0.101 M.

• The concentration of \(\mathrm{NH}_4^+\) is 0.099 M.

Substitute into:\[ pH = 9.25 + \log\left(\frac{0.101}{0.099}\right) \]Simplifying gives:\[ pH = 9.25 + 0.009 \]\[ pH = 9.259 \]This shows a slight increase in pH, demonstrating the buffer's ability to dampen drastic pH changes.

Acid-Base Equilibrium

Understanding acid-base equilibrium is crucial for working with buffer solutions. In any buffer solution, there exists an equilibrium between the weak acid and its conjugate base. This dynamic balance allows buffers to resist drastic pH changes when small amounts of strong acids or bases are added.
This is because:

• Adding a base shifts the equilibrium towards forming more of the acid form, reducing pH increase.
• Adding an acid shifts the equilibrium towards forming more of the base form, reducing pH decrease.

The equilibrium can be represented by the reversible equation:\[ \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \]When you introduce KOH into the buffer solution, the hydroxide ions \(\mathrm{OH}^-\) react with ammonium ions \(\mathrm{NH}_4^+\), converting them to ammonia \(\mathrm{NH}_3\).

This highlights how buffers maintain equilibrium by adjusting ion concentrations and hence stabilizing the pH, ensuring that any change remains minimal and within a predictable range.