Question

What is the pH of the solution that results from adding \(25.0 \mathrm{mL}\) of \(0.12 \mathrm{M} \mathrm{HCl}\) to \(25.0 \mathrm{mL}\) of \(0.43 \mathrm{M} \mathrm{NH}_{3} ?\)

Short answer

The pH of the solution is approximately 9.66.

Step-by-step solution

Write the Reaction

The chemical reaction between hydrochloric acid (HCl) and ammonia (NH₃) is: \[ \mathrm{NH}_3 + \mathrm{HCl} \rightarrow \mathrm{NH}_4^+ + \mathrm{Cl}^- \] Ammonia reacts with hydrochloric acid to form ammonium ions (\(\mathrm{NH}_4^+\)) and chloride ions (\(\mathrm{Cl}^-\)).

Calculate Moles of Each Reactant

Calculate the moles of \(\mathrm{HCl}\) and \(\mathrm{NH}_3\): - Moles of \(\mathrm{HCl}\): \( 0.12 \, \mathrm{M} \times 0.025 \, \mathrm{L} = 0.003 \, \text{mol} \)- Moles of \(\mathrm{NH}_3\): \( 0.43 \, \mathrm{M} \times 0.025 \, \mathrm{L} = 0.01075 \, \text{mol} \)

Determine Limiting Reactant

Since 0.003 mol of \(\mathrm{HCl}\) is less than 0.01075 mol of \(\mathrm{NH}_3\), \(\mathrm{HCl}\) is the limiting reactant. This means all \(\mathrm{HCl}\) will react.

Find Excess Reactant

After the reaction, the moles of \(\mathrm{NH}_3\) left unreacted are: \[ 0.01075 \, \text{mol} - 0.003 \, \text{mol} = 0.00775 \, \text{mol of NH}_3 \] Also, 0.003 mol of \(\mathrm{NH}_4^+\) is formed.

Calculate Total Volume of Solution

The total volume of the mixed solution is:\[ 25.0 \, \text{mL} + 25.0 \, \text{mL} = 50.0 \, \text{mL} = 0.0500 \, \text{L} \]

Calculate Concentration of \(\mathrm{NH}_3\) and \(\mathrm{NH}_4^+\)

Find the concentrations in the final solution:- \([\mathrm{NH}_3] = \frac{0.00775 \, \text{mol}}{0.0500 \, \text{L}} = 0.155 \, \mathrm{M}\)- \([\mathrm{NH}_4^+] = \frac{0.003 \, \text{mol}}{0.0500 \, \text{L}} = 0.060 \, \mathrm{M}\)

Calculate pH Using \(\mathrm{NH}_3\) and \(\mathrm{NH}_4^+\)

The \(\mathrm{NH}_4^+/\mathrm{NH}_3\) buffer system can be used. Apply the Henderson-Hasselbalch equation:\[ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right) \]Given \(\mathrm{pK}_a = 9.25\) for \(\mathrm{NH}_4^+\), calculate:\[ \mathrm{pH} = 9.25 + \log\left(\frac{0.155}{0.060}\right) \approx 9.25 + 0.41 \approx 9.66 \]

Key concepts

Acid-Base Reaction

An acid-base reaction is a chemical process that involves the transfer of protons (H⁺ ions) between substances. In this exercise, we analyze the reaction between hydrochloric acid (HCl), a strong acid, and ammonia (NH₃), a weak base.
When HCl is added to NH₃, the reaction produces ammonium ions (NH₄⁺) and chloride ions (Cl⁻). This can be represented by the equation: NH₃ + HCl → NH₄⁺ + Cl⁻.
The reaction between a strong acid and a weak base helps in understanding how ions and molecules interact in solution. This lays the groundwork for further understanding of buffer systems, which play a key role in pH stabilization in various biological and chemical environments.

Limiting Reactant

The concept of a limiting reactant plays a crucial role in calculating the pH of solutions resulting from acid-base reactions. Here, the limiting reactant determines how much product can be formed.
In our exercise, the moles of HCl are fewer than the moles of NH₃, making HCl the limiting reactant. This is significant because only 0.003 mol of NH₃ can react due to the limited amount of HCl.
Understanding which reactant is limiting helps determine the amount of product formed and the amount of leftover excess reactant. For this reaction, it results in all HCl being consumed and leaves some NH₃ unreacted. This remaining NH₃ will further influence the concentration values used in buffer solution calculations.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool for determining the pH of buffer solutions. This equation relates the pH of a solution to the pKa and the ratio of the concentrations of an acidic form of a compound and its conjugate base.
In mathematical terms, it is expressed as: \[\mathrm{pH} = \mathrm{pK_a} + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]
For the ammonia and ammonium system, where NH₄⁺ is the acidic form and NH₃ is the base, this equation allows us to calculate the pH.
By substituting the concentrations, pK_a, and base-over-acid ratio into this equation, the pH derived in our example is approximately 9.66. This calculation illustrates the buffering capacity and stability conferred by the ratio of NH₃ and NH₄⁺.

Buffer Solution

A buffer solution is a special aqueous solution that resists changes in pH when small amounts of an acid or a base are added. It is composed of a weak acid and its conjugate base or a weak base and its conjugate acid.
The reaction we analyzed forms an ammonia and ammonium buffer solution. When NH₃ reacts with HCl, NH₄⁺ is formed, and together they create a buffer.

• NH₃ is the weak base.
• NH₄⁺ is its conjugate acid.

This buffer system stabilizes the pH of the solution despite potential changes from additional acids or bases. The stable pH is crucial in many biological and chemical environments where enzymes and reactions are sensitive to pH changes. Understanding buffer solutions is key to grasping how biological systems maintain homeostasis and chemical reactions proceed efficiently.