To calculate the pH of this buffer solution, we can use the Henderson-Hasselbalch equation for buffer solutions, given by:

[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

where:

• $\text{p}K_a = -\log(K_a)$
• $[\text{A}^-]$ is the concentration of the conjugate base (sodium acetate, NaC₂H₃O₂)
• $[\text{HA}]$ is the concentration of the weak acid (acetic acid, HC₂H₃O₂)

Given Data:

• Volume of solution, $V = 1.00 \, \text{L}$
• Moles of acetic acid, $\text{HA} = 0.153 \, \text{mol}$
• Moles of sodium acetate, $\text{A}^- = 0.194 \, \text{mol}$
• $K_a$ of acetic acid = $1.8 \times 10^{-5}$

Step 1: Calculate $\text{p}K_a$

$\text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74$

Step 2: Calculate concentrations of HA and A⁻

Since the solution volume is 1.00 L, the concentrations are simply the moles per liter: $[\text{HA}] = \frac{0.153 \, \text{mol}}{1.00 \, \text{L}} = 0.153 \, \text{M}$ $[\text{A}^-] = \frac{0.194 \, \text{mol}}{1.00 \, \text{L}} = 0.194 \, \text{M}$

Step 3: Apply the Henderson-Hasselbalch equation

$\text{pH} = 4.74 + \log\left(\frac{0.194}{0.153}\right)$ Calculating the logarithmic term: $\log\left(\frac{0.194}{0.153}\right) \approx 0.110$ Thus, $\text{pH} = 4.74 + 0.110 = 4.85$

Final Answer:

The pH of the buffer solution is approximately 4.85.

Would you like a detailed explanation of each step or have any further questions?

Here are five related questions to explore further:

1. How does the pH of a buffer change if more acid is added to the solution?
2. What would happen to the pH if the concentration of sodium acetate was decreased?
3. How would the pH calculation differ if the buffer was diluted?
4. Can this buffer resist changes in pH if a strong base is added? How would you calculate that?
5. What is the effective pH range for this buffer solution?

Tip: Buffers work best within a range of ±1 pH unit from the pKa of the weak acid.