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

The concentration of $\mathrm{NH}_{4} \mathrm{Cl}$ needed to buffer a $0.52 \mathrm{M}$ $\mathrm{NH}_{3}$ solution at pH 9.00 is \(9.36 \times 10^{-4}\, M\).

Step-by-step solution

Find the pOH and the OH⁻ concentration

Since we are given the pH, we will first find the pOH using the relationship: \[pH + pOH = pK_w = 14\] where \(pK_w\) is the ionic product of water. Now we can calculate the pOH: \[pOH = 14 - pH = 14 - 9 = 5\] Now that we have the pOH, we can calculate the \(OH^-\) concentration using the formula: \[OH^- = 10^{-pOH}\] \[OH^- = 10^{-5}\]

Set up the Kb expression

For NH\(_3\) the reaction equation is: \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\] Given the base dissociation constant \(K_b\) for NH\(_3\): \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\] We need to find the concentration of \(NH_4^+\).

Use the Kb expression to find the concentration of NH4⁺

Using the equation from step 2, we can substitute the values of \(K_b\), \([OH^-]\), and \([NH_3]\). \[ K_b = 1.8 \times 10^{-5} = \frac{[NH_4^+](10^{-5})}{0.52}\] Next, we need to solve for the \([NH_4^+]\): \[ [NH_4^+] = 1.8 \times 10^{-5} \times \frac{0.52}{10^{-5}}\] \[ [NH_4^+] = 9.36 \times 10^{-4} M\]

Determine the concentration of NH4Cl needed

The concentration of NH\(_4\)Cl is equal to the concentration of the NH\(_4^+\) ion released in the solution, so the concentration of NH\(_4\)Cl needed to buffer the 0.52 M NH\(_3\) solution at pH 9.00 is: \[ [NH_4Cl] = [NH_4^+] = 9.36 \times 10^{-4} M\]

Key concepts

pH calculation

To understand how to calculate pH, it's essential to grasp the relationship between pH, pOH, and the ionic product of water. The basic formula that connects these values is:

• \( pH + pOH = 14 \)

This equation shows that at 25°C, the sum of pH and pOH is always equal to 14. If you know the pH, you can subtract it from 14 to find the pOH, as demonstrated in the buffer question where the given pH is 9.
Thus, \( pOH = 14 - 9 = 5 \).
The OH⁻ concentration can then be calculated using:

• \( [OH^-] = 10^{-pOH} \)

So, plug in the values: \( [OH^-] = 10^{-5} \).This basic understanding of pH and pOH calculations is foundational in equilibrium chemistry.
Having these skills helps you dive deeper into buffer solutions and equilibrium equations.

base dissociation constant

The base dissociation constant, \( K_b \), is a key player in equilibrium chemistry. It helps us understand how bases dissociate in water.
For ammonia (\( NH_3 \)), the dissociation in water is:

• \( NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \)

The \( K_b \) expression for this equilibrium is:

• \( K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \)

This formula shows the relationship between the concentrations of the ammonia, ammonium ions, and hydroxide ions.
Given that \( K_b \) for ammonia is \( 1.8 \times 10^{-5} \), we use this to find unknown concentrations like \([NH_4^+]\).
This process helps understand the strength of a base and its ability to dissociate.
By resolving the equilibrium expression, you can discover the necessary quantities to achieve a certain pH buffer, which is crucial in analytical chemistry.

equilibrium chemistry

Equilibrium chemistry revolves around the idea that chemical reactions can reach a state where the reactants and products remain constant over time.
This is known as dynamic equilibrium, where the forward and reverse reactions occur at the same rate. In the context of the given problem:

• The reaction: \( NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \) exemplifies such a balance.

In buffer solutions, equilibrium chemistry is incredibly important. Buffers resist changes in pH when small amounts of acid or base are added.
When solving problems involving buffer solutions, such as the concentration of \( NH_4Cl \) necessary to achieve a specific pH, understanding equilibrium chemistry allows us to establish and solve the equations that dictate these balances.
By leveraging the concept of \( K_b \) and equilibrium constants, we can precisely calculate the amounts required for balanced reactions.
Equilibrium chemistry is foundational in predicting and manipulating the behavior of chemical reactions in various environments.