Question

Calculation of pH from Molar Concentrations The \(\mathrm{p} K_{a}\) of \(\mathrm{NH}_{4}^{+} / \mathrm{NH}_{3}\) is 9.25. Calculate the \(\mathrm{pH}\) of a solution containing \(0.12 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) and \(0.03 \mathrm{NaOH}\).

Short answer

The pH of the solution is 8.77.

Step-by-step solution

Understand the Chemical Reaction

In this solution, we have \(0.12\,\text{M}\, \text{NH}_4\text{Cl}\) and \(0.03\,\text{M}\, \text{NaOH}\),which react as \(\text{NH}_4^+\) reacts with the added hydroxide ions (from \(\text{NaOH}\)) to form \(\text{NH}_3\) and water:\[\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}\]

Calculate Moles of \\ \text{NH}_4^+ and \\ \text{OH}^-

Calculate the moles of each reactant:- \(\text{Moles of } \text{NH}_4^+ = 0.12 \,\text{mol/L}\)- \(\text{Moles of } \text{OH}^- = 0.03 \,\text{mol/L}\)

Determine Limiting Reactant

The \(\text{OH}^-\) is the limiting reactant because it is present in lower concentration than \(\text{NH}_4^+\). All \(\text{OH}^-\) will react to form \(\text{NH}_3\), and the decrease in \(\text{NH}_4^+\) will be equal to the initial \(\text{OH}^-\) concentration.

Calculate Concentrations after Reaction

After neutralization, - \([\text{NH}_4^+] = 0.12 - 0.03 = 0.09\,\text{M}\)- \([\text{NH}_3] = 0.03\,\text{M}\) (generated from the reaction)Now we have an \(\text{NH}_4^+/\text{NH}_3\) buffer.

Use Henderson-Hasselbalch Equation for Buffer pH

We use the Henderson-Hasselbalch equation to find the pH:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{NH}_3]}{[\text{NH}_4^+]}\right)\]Substitute the values:\[ \text{pH} = 9.25 + \log\left(\frac{0.03}{0.09}\right)\]\[ \text{pH} = 9.25 + \log(0.33) = 9.25 - 0.48 = 8.77\]

Verify pH Value

The solution is slightly basic as expected for a mixture containing ammonia (\(\text{NH}_3\)) and the conjugate acid (\(\text{NH}_4^+\)). The calculated pH of 8.77 aligns with the conditions of the buffer, confirming the calculation's accuracy.

Key concepts

Buffer Solution

A buffer solution is a special kind of solution that resists changes in its pH when small amounts of acid or base are added. Buffer solutions contain a weak acid and its conjugate base, or a weak base and its conjugate acid. In this particular exercise, we have a mixture containing

• Ammonium ions (\(\mathrm{NH}_4^+\)), which act as the weak acid
• Ammonia (\(\mathrm{NH}_3\)), which acts as the conjugate base

These components create what is known as an ammonium buffer system.

Buffers are crucial in maintaining the stability of pH in various environments, such as biological systems and chemical reactions. The special characteristic of buffers is their ability to "neutralize" or minimize the impact of added acids or bases, maintaining the pH within a narrow range.

In our exercise, after the reaction between ammonium ions and hydroxide ions, the leftover ammonia (\(\mathrm{NH}_3\)) and ammonium ions (\(\mathrm{NH}_4^+\)) form a buffer solution that helps in stabilizing the pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable formula used to estimate the pH of a buffer solution. It is particularly helpful because it relates the pH of the solution directly to the concentration of the acid and its conjugate base in the buffer.

The formula is:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right)\]In this case, the base is ammonia (\(\mathrm{NH}_3\)) and the acid is ammonium ions (\(\mathrm{NH}_4^+\)). The value of \(\text{pK}_a\) for the \(\mathrm{NH}_4^+/\mathrm{NH}_3\) system is given as 9.25.

To apply the equation, you plug in the concentrations of the base and acid in the buffer system after the reaction takes place. This helps in calculating the pH effectively. It's a simplified way to find pH without having to calculate the exact concentrations of all ions in solution, which can be more complex. The equation assumes that the changes in volume are negligible and that the "ideal" approximation of activities apply.

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between acids and bases in a solution. It is a central concept in understanding buffer solutions, as well as many chemical and biological processes. Equilibrium is reached in a reversible reaction where the forward and reverse reaction rates are equal, resulting in a stable concentration of reactants and products.

In the context of this exercise, we consider the equilibrium between ammonia (\(\text{NH}_3\)) and ammonium ions (\(\text{NH}_4^+\)). Ammonium ion acts as a weak acid and can donate a proton, while ammonia acts as a weak base and can accept a proton. This balance is key to forming a buffer.

The addition of hydroxide ions (\(\text{OH}^-\)) to the solution initially shifts the equilibrium by reacting with the ammonium ions. This reaction converts some \(\text{NH}_4^+\) to \(\text{NH}_3\) until the \(\text{OH}^-\) is consumed. This process illustrates how buffer systems utilize the reversible reactions to maintain a stable pH even as components are added.