- The Brønsted-Lowry theory defines an acid as a proton donor and a base as a proton acceptor.
- Set up an ICE table with an initial concentration of \(0.335\text{ M}\) for \(NH_3\).
- Use the base dissociation constant \(K_b = 1.8 \times 10^{-5}\) to find the equilibrium concentration of \(OH^-\) as \(x \approx 2.456 \times 10^{-3}\text{ M}\).
- Calculate pOH as \(\approx 2.61\) and the final pH as \(\boxed{11.39}\).
Explanation
- Defining the Brønsted-Lowry Theory
The Brønsted-Lowry acid-base theory provides a comprehensive definition of acids and bases based on proton transfer. A Brønsted-Lowry acid is defined as a substance that can donate a proton (\(H^+\)), while a Brønsted-Lowry base is a substance that can accept a proton (\(H^+\)). In the context of the reaction of ammonia (\(NH_3\)) with water, ammonia acts as a base by accepting a proton from the water molecule:
\[NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)\]
- Completing the ICE Chart
To determine the equilibrium concentration, we set up an Initial-Change-Equilibrium (ICE) table. We start with an initial concentration of \(NH_3\) equal to \(0.335\text{ M}\), while the concentrations of \(NH_4^+\) and \(OH^-\) are negligible initially. Let \(x\) represent the change in concentration for each species as the system moves towards equilibrium:
\[ \begin{array}{|l|c|c|c|} \ & \text{NH}_3 & \text{NH}_4^+ & \text{OH}^- \\ \text{Initial} & 0.335 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 0.335 - x & x & x \\ \end{array} \]
- Calculating Hydroxide Ion Concentration
Using the base dissociation constant expression \(K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\), we substitute the equilibrium values from the ICE chart:
\[K_b = \frac{x \cdot x}{0.335 - x} = \frac{x^2}{0.335 - x}\]
Given that \(K_b = 1.8 \times 10^{-5}\) is very small, we can assume that \(x\) is much smaller than \(0.335\), allowing the approximation \(0.335 - x \approx 0.335\). The expression simplifies to:
\[1.8 \times 10^{-5} \approx \frac{x^2}{0.335}\]
Solving for \(x^2\):
\[x^2 \approx 1.8 \times 10^{-5} \times 0.335 = 6.03 \times 10^{-6}\]
Taking the square root:
\[x = \sqrt{6.03 \times 10^{-6}} \approx 2.456 \times 10^{-3}\text{ M}\]
This \(x\) value represents the concentration of hydroxide ions, \([OH^-]\).
- Final Calculation of pH
Now we calculate the pOH using the formula \(\text{pOH} = -\log_{10}[OH^-]\):
\[\text{pOH} = -\log_{10}(2.456 \times 10^{-3}) \approx 2.61\]
Finally, the pH of the solution is determined using the relationship \(\text{pH} + \text{pOH} = 14.00\):
\[\text{pH} = 14.00 - 2.61 = 11.39\]
The pH of the \(0.335\text{ M}\) ammonia solution is approximately \(\boxed{11.39}\).
Examples
The concept of pH is essential in environmental science and medicine. For instance, when monitoring the water quality of a local lake, chemists measure the pH to ensure it remains stable for aquatic life. Similarly, in your own body, the blood's pH must remain within a very narrow range (around 7.4) to function correctly. Calculating the pH of a substance like ammonia helps us understand how basic or acidic it is, which is crucial for safety in laboratories and manufacturing processes.