Problem 125 - General Chemistry

Question

A solution made up of \(1.0 \mathrm{M} \mathrm{NH}_{3}\) and \(0.50 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) has a pH of 9.26 . a. Write the net ionic equation that represents the reaction of this solution with a strong acid. b. Write the net ionic equation that represents the reaction of this solution with a strong base. c. To \(100 . \mathrm{mL}\) of this solution, \(10.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added. How many moles of \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}\) are present in the reaction system before and after the addition of the \(\mathrm{HCl}\) ? What is the \(\mathrm{pH}\) of the resulting solution? d. Why did the \(\mathrm{pH}\) change only slightly upon the addition of \(\mathrm{HCl} ?\)

Step-by-Step Solution

Answer

The net ionic equation with a strong acid is \( \text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+ \). The net ionic equation with a strong base is \( \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \). After adding \( \text{HCl} \), the pH changes slightly to 9.15 due to buffering.

Step 1: Identify the net ionic equation with a strong acid

When a strong acid is added to this solution, the ammonia (\( \text{NH}_3 \)) will react with the hydrogen ions (\( \text{H}^+ \)) from the acid. The net ionic equation is:\[\text{NH}_3 (aq) + \text{H}^+ (aq) \rightarrow \text{NH}_4^+ (aq)\]This equation shows that ammonia accepts a proton from the strong acid, forming ammonium ions.

Step 2: Identify the net ionic equation with a strong base

When a strong base is added, the ammonium ions (\( \text{NH}_4^+ \)) will react with hydroxide ions (\( \text{OH}^- \)). The net ionic equation is:\[\text{NH}_4^+ (aq) + \text{OH}^- (aq) \rightarrow \text{NH}_3 (aq) + \text{H}_2\text{O} (l)\]This equation shows that ammonium ions donate a proton to hydroxide ions, forming ammonia and water.

Step 3: Calculate initial moles of NH3 and NH4+

The initial volume of solution is 100 mL, which is 0.1 L. Calculate the initial moles of ammonia and ammonium sulfate.\[\text{For } \text{NH}_3: (1.0 \text{ M solution} \times 0.1 \text{ L}) = 0.1 \text{ mol}\]Ammonium sulfate contributes two \( \text{NH}_4^+ \) ions per formula unit, thus:\[\text{For } \text{NH}_4^+: (0.50 \text{ M} \times 0.1 \text{ L} \times 2) = 0.10 \text{ mol}\]Initially, there are 0.1 mol of \( \text{NH}_3 \) and 0.1 mol of \( \text{NH}_4^+ \).

Step 4: Add moles of HCl and calculate change in NH3 and NH4+

Adding 10 mL of 1.0 M HCl gives \[10 \text{ mL} \times 1.0 \text{ M} = 0.010 \text{ mol HCl}\]HCl fully dissociates and reacts with \( \text{NH}_3 \), reducing its moles and increasing \( \text{NH}_4^+ \) moles by the same amount:\[\text{NH}_3: 0.1 - 0.01 = 0.09 \text{ mol}\]\[\text{NH}_4^+: 0.1 + 0.01 = 0.11 \text{ mol}\]After addition, \( \text{NH}_3 \) decreases to 0.09 mol and \( \text{NH}_4^+ \) increases to 0.11 mol.

Step 5: Calculate new pH of the solution

Using the Henderson-Hasselbalch equation:\[pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]Given \( pK_b \) for \( \text{NH}_3 \) is 4.75, we find \( pK_a = 14 - 4.75 = 9.25 \). Calculate the pH:\[pH = 9.25 + \log \left( \frac{0.09}{0.11} \right)\]Which approximately equals 9.15. The pH changed from 9.26 to 9.15.

Step 6: Explain slight pH change

The pH changed only slightly because this is a buffer solution. The presence of both \( \text{NH}_3 \) and \( \text{NH}_4^+ \) allows the solution to resist changes in pH upon the addition of small amounts of strong acid or base, maintaining the pH near its initial value.

Key Concepts

Understanding Net Ionic Equations

Net ionic equations are essential tools in chemistry to simplify reaction descriptions by showing only the particles that participate directly in the reaction. In particular, when examining buffer solutions like in our exercise, net ionic equations help illustrate the chemical behavior in a clear and concise manner.

In a buffer solution consisting of ammonia (\( \text{NH}_3 \)) and ammonium sulfate, adding a strong acid such as hydrochloric acid (HCl) results in a net ionic equation. Here, \( \text{NH}_3 \) accepts protons from the dissociated \( \text{H}^+ \) ions, forming ammonium ions (\( \text{NH}_4^+ \)). The simple reaction can be represented as:

• \( \text{NH}_3 (aq) + \text{H}^+ (aq) \rightarrow \text{NH}_4^+ (aq) \)

Conversely, when a strong base is added, the net ionic equation shows the \( \text{NH}_4^+ \) donating a proton to form \( \text{NH}_3 \) and water, as follows:

• \( \text{NH}_4^+ (aq) + \text{OH}^- (aq) \rightarrow \text{NH}_3 (aq) + \text{H}_2\text{O} (l) \)

These reactions display the dynamic exchange of protons typical in buffer systems.

Exploring the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital formula for understanding and calculating the pH of buffer solutions. This equation links the pH level of a solution to the concentration of an acid and its corresponding base.

For ammonium and ammonia, this equation makes calculating pH straightforward. The formula is:

• \[pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]

In our scenario, ammonia acts as the base, and ammonium ions act as the acid. By knowing the \( pK_a \) value, derived from the \( pK_b \) of ammonia and the ion product of water, we simplify the calculation process. Given a \( pK_b \) of 4.75 for ammonia, we find \( pK_a = 14 - 4.75 = 9.25 \).

The equation demonstrates the pivotal role of the

• ratio between ammonia and ammonium concentration
• in establishing the pH of the buffer system.

This approach is crucial for predicting pH changes upon the addition of strong acids or bases to buffer solutions.

The Basics of pH Calculation in Buffer Solutions

Calculating the pH of a buffer solution involves understanding how the solution's components interact when introduced to excess acid or base. Consider the effect of adding hydrochloric acid (HCl) to our buffer.

Initially, in the solution, we calculate the moles of \( \text{NH}_3 \) and \( \text{NH}_4^+ \) using their concentrations and the volume of the solution. Adding HCl alters these concentrations as it reacts with the \( \text{NH}_3 \), reducing its quantity while increasing \( \text{NH}_4^+ \).

• Before addition: \( \text{NH}_3 = 0.1 \text{ mol} \) and \( \text{NH}_4^+ = 0.1 \text{ mol} \)
• After adding 0.01 mol HCl: \( \text{NH}_3 \) becomes 0.09 mol, and \( \text{NH}_4^+ \) turns into 0.11 mol.

Using the Henderson-Hasselbalch equation, we assess how pH shifts slightly as the concentrations change, reaffirming the buffer's capacity to maintain pH stability. In calculating, we find that the pH diminishes slightly from 9.26 to 9.15.

This minor pH drop illustrates the robust resistance of buffer solutions to drastic pH changes, even with the addition of acidic or basic substances.