Problem 140 - General Chemistry

Question

The \(K_{b}\) for \(\mathrm{NH}_{3}\) is \(1.8 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) of a buffer solution made by mixing \(65.1 \mathrm{~mL}\) of \(0.142 \mathrm{M}\) \(\mathrm{NH}_{3}\) with \(38.0 \mathrm{~mL}\) of \(0.172 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) at \(25^{\circ} \mathrm{C}\). Assume that the volumes of the solutions are additive.

Step-by-Step Solution

Answer

The pH of the buffer solution is approximately 9.41.

Step 1: Identify the Problem Type

This problem involves calculating the pH of a buffer solution made from weak base ammonia, \(\mathrm{NH}_3\), and its conjugate acid, ammonium chloride, \(\mathrm{NH_4Cl}\). This involves using the buffer equation, which requires the \(K_b\) to find \(K_a\) and then the \(pK_a\).

Step 2: Calculate the Number of Moles

First, calculate the moles of \(\mathrm{NH_3}\) and \(\mathrm{NH_4Cl}\).- Moles of \(\mathrm{NH_3} = 0.142 \, \mathrm{M} \times 0.0651 \, \mathrm{L} = 0.0092442 \, \mathrm{moles}\)- Moles of \(\mathrm{NH_4Cl} = 0.172 \, \mathrm{M} \times 0.0380 \, \mathrm{L} = 0.006536 \, \mathrm{moles}\)

Step 3: Calculate the Concentrations

Since the volumes are additive, the total volume of the buffer solution is \(65.1 \, \mathrm{mL} + 38.0 \, \mathrm{mL} = 103.1 \, \mathrm{mL} = 0.1031 \, \mathrm{L}\).- Concentration of \(\mathrm{NH_3} = \frac{0.0092442}{0.1031} = 0.08966 \, \mathrm{M}\)- Concentration of \(\mathrm{NH_4Cl} = \frac{0.006536}{0.1031} = 0.06341 \, \mathrm{M}\)

Step 4: Calculate the \(K_a\) of Ammonium Ion

We use the relationship between \(K_a\) and \(K_b\): \(K_w = K_a \cdot K_b\), where \(K_w = 1.0 \times 10^{-14}\).\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]

Step 5: Find the \(pK_a\)

The \(pK_a\) is the negative logarithm of \(K_a\).\[ pK_a = -\log(5.56 \times 10^{-10}) = 9.255 \]

Step 6: Apply the Henderson-Hasselbalch Equation

Using the concentrations and \(pK_a\), apply the Henderson-Hasselbalch equation:\[ \mathrm{pH} = pK_a + \log \left(\frac{[\mathrm{base}]}{[\mathrm{acid}]}\right) \]\[ \mathrm{pH} = 9.255 + \log \left(\frac{0.08966}{0.06341}\right) = 9.255 + \log(1.413) \]\[ \mathrm{pH} = 9.255 + 0.150 = 9.405 \]

Step 7: Conclusion

The calculated pH of the buffer solution is approximately 9.41.

Key Concepts

Henderson-Hasselbalch Equation

Buffer solutions are crucial in maintaining stable pH levels in various chemical and biological systems. One effective way to calculate the pH of such a solution is by using the Henderson-Hasselbalch equation. This equation simplifies the pH calculation for a buffer solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid. The formula is:\[ \mathrm{pH} = pK_a + \log \left(\frac{[\mathrm{base}]}{[\mathrm{acid}]}\right) \]In the context of the provided problem, this equation helps us determine the pH of the buffer solution consisting of ammonia (weak base) and ammonium chloride (conjugate acid). By calculating the concentrations of the base and the acid in the solution, and using the previously calculated \(pK_a\), we can efficiently find the pH. The equation assumes good mixing and that the volume is additive, which simplifies handling different concentrations.

• It provides a practical way to explore how changes in concentration affect pH.
• Useful in predicting buffer capacity.

Understanding this equation is vital for chemists and biochemists who work with solutions where pH stability is essential.

Weak Base and Conjugate Acid

In chemistry, a weak base is a base that does not completely dissociate in solutions, meaning not all of its molecules donate hydroxide ions \(\mathrm{OH^-}\) to the solution. This partial dissociation creates an equilibrium between the base and its odd ions in the solution, leading to relatively mild changes in pH when acids or bases are added. Ammonia (\(\mathrm{NH_3}\)) is a classic example.In a buffer solution, the role of the weak base and its counterpart, the conjugate acid, is crucial. The conjugate acid of \(\mathrm{NH_3}\) is ammonium \(\mathrm{NH_4^+}\), as found in ammonium chloride. When \(\mathrm{NH_3}\) accepts a proton (\(\mathrm{H^+}\)), it becomes \(\mathrm{NH_4^+}\). This reaction can be reversed when \(\mathrm{NH_4^+}\) donates a proton back, exemplifying buffer action:- This pair resists drastic changes in pH with the common ion effect.- Adding acid shifts the balance, leading more \(\mathrm{NH_3}\) to form \(\mathrm{NH_4^+}\).- Adding base increases \(\mathrm{NH_3}\) by removing \(\mathrm{H^+}\) from \(\mathrm{NH_4^+}\).Thus, the balance between ammonia and ammonium ions maintains the pH, demonstrating the power of weak base-conjugate acid pairs in buffer solutions.

Chemical Equilibrium Constants

Chemical equilibrium constants play a vital role in understanding the behavior of systems in equilibrium and determining concentrations of different species. For buffer solutions involving weak acids and bases, two kinds of equilibrium constants are crucial: the base dissociation constant \(K_b\) and the acid dissociation constant \(K_a\).Why do we care about \(K_b\) and \(K_a\)? These constants tell us about the strength of the acid or base involved. Specifically, they provide insights into how readily an acid donates protons or how easily a base accepts them. For ammonia:- Given \(K_b = 1.8 \times 10^{-5}\) for \(\mathrm{NH_3}\), it indicates ammonia does not fully dissociate in water.- Consequently, \(K_a\) can be determined through the water dissociation constant \(K_w (1.0 \times 10^{-14})\) via \(K_a = \frac{K_w}{K_b}\).These constants are crucial for implementing the Henderson-Hasselbalch equation. They allow for accurate pH predictions in buffers, making them indispensable tools in chemistry. Understanding how these constants interrelate gives you a powerful handle on predicting the behavior of solutions under different circumstances.