Question

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}} \text { for } \mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

Short answer

A concentration of approximately \(0.35 \, \text{M}\) \(\mathrm{NH}_{4}\mathrm{Cl}\) is necessary to buffer a \(0.52-\text{M}\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH} = 9.00\).

Step-by-step solution

Find pOH

Given \(\mathrm{pH}=9.00\), we can calculate the \(\mathrm{pOH}\) using the following equation: \(\mathrm{pH}+\mathrm{pOH}=14\) Solve for \(\mathrm{pOH}\): \(\mathrm{pOH}=14-\mathrm{pH}=14-9.00=5.00\)

Calculate pK\(_b\)

Using the given \(K_\mathrm{b}=1.8 \times 10^{-5}\), we find the \(pK_b\) using the formula below: \(pK_\mathrm{b}=-\log K_\mathrm{b}\) \(pK_\mathrm{b}=-\log (1.8 \times 10^{-5}) = 4.74\)

Apply the Henderson-Hasselbalch equation

Now, apply the Henderson-Hasselbalch equation for buffer solutions: \(pOH = pK_b + \log \frac{[\mathrm{A}^-]}{[\mathrm{HB}^+]}\) Insert the values we found for \(\mathrm{pOH}\) and \(pK_b\) and the concentration of \(\mathrm{NH}_3\) into the equation: \(5.00 = 4.74 + \log \frac{0.52}{[\mathrm{NH}_4^+]}\)

Solve for \(\mathrm{[NH}_4^{+}\mathrm{]}\)

To find the concentration of \(\mathrm{NH}_{4}\mathrm{Cl}\), which corresponds to the concentration \([\mathrm{NH}_4^+]\), we will solve the equation: \(5.00 - 4.74 = \log \frac{0.52}{[\mathrm{NH}_4^+]}\) \(0.26 = \log \frac{0.52}{[\mathrm{NH}_4^+]}\) Now, raise both sides with the base \(10\) to eliminate the log function: \(10^{0.26} = \frac{0.52}{[\mathrm{NH}_4^+]}\) Now, solve for \([\mathrm{NH}_4^+]\): \([\mathrm{NH}_4^+] = \frac{0.52}{10^{0.26}}\) \([\mathrm{NH}_4^+] \approx 0.35 \, \text{M}\) Therefore, a concentration of approximately \(0.35 \, \text{M}\) \(\mathrm{NH}_{4}\mathrm{Cl}\) is necessary to buffer a \(0.52-\text{M}\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH} = 9.00\).

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is central to understanding buffer solutions. It provides a direct relationship between the pH (or pOH) of a solution and the concentrations of a weak acid and its conjugate base or a weak base and its conjugate acid.

Particularly, the equation is expressed as: \[\text{pH} = pK_a + \log \frac{[\text{A}^-]}{[\text{HA}]}\]For weak bases and their conjugate acids, it is written in terms of pOH and pKb: \[\text{pOH} = pK_b + \log \frac{[\text{B}]}{[\text{BH}^+]}\]
Where

• \(pK_a\) and \(pK_b\) are the negative logarithms of the acid dissociation constant and base dissociation constant, respectively,
• \([\text{A}^-]\) and \([\text{B}]\) represent the concentration of the conjugate base of the weak acid and the concentration of the weak base, respectively,
• \([\text{HA}]\) and \([\text{BH}^+]\) stand for the concentration of the undissociated weak acid and the concentration of the conjugate acid of the weak base.

When preparing a buffer solution, you adjust the ratio of \([\text{A}^-]/[\text{HA}]\) or \([\text{B}]/[\text{BH}^+]\) so that the pH or pOH of the solution will resist changes upon the addition of small amounts of acid or base.

pH and pOH

pH and pOH are both measures of the acidity and basicity of a solution. The pH scale is used to determine how acidic or basic a water-based solution is, with lower values indicating higher acidity and higher values indicating higher basicity. The pOH scale is similar, but it specifically indicates the concentration of hydroxide ions \(\text{OH}^-\).

The relationship between pH and pOH can be stated as: \[\text{pH} + \text{pOH} = 14\]This is because the product of the hydronium ions \(\text{H}_3\text{O}^+\) and hydroxide ions \(\text{OH}^-\) concentrations are constant at a given temperature, for pure water or any aqueous solution. To simplify the process, finding the pOH can be as straightforward as subtracting the pH from 14, as seen in our exercise, leading us to further understand the properties of the buffer solution.

Weak Base and Conjugate Acid

A weak base is a base that does not completely dissociate into its ions in a solution. Ammonia \(\text{NH}_3\) is a common example of a weak base. Its dissociation in water is incomplete, and it exists in equilibrium with its ions and undissociated molecules.

When a weak base accepts a proton \(\text{H}^+\), it forms its conjugate acid. For ammonia, the conjugate acid is the ammonium ion \(\text{NH}_4^+\). This relationship between a weak base and its conjugate acid is critical in buffer systems. Specifically:

• The weak base helps neutralize added acid,
• The conjugate acid neutralizes added base,
• The ratio of weak base to conjugate acid determines the buffer's pH.

In the exercise, the presence of \(\text{NH}_4\text{Cl}\), which provides the \(\text{NH}_4^+\), is necessary because it is the source of the conjugate acid for the weak base \(\text{NH}_3\). The ratio of \(\text{NH}_3\) to \(\text{NH}_4^+\) will govern the buffer's pH, enabling it to resist changes in pH when acids or bases are added to the solution.