To calculate the percent ionization of a 0.15 M benzoic acid solution in the presence of 0.10 M sodium benzoate, we can use the Henderson-Hasselbalch equation and the given $K_a$.

Step 1: Calculate the pH using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by:

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

Where:

• $[\text{A}^-]$ is the concentration of the conjugate base (sodium benzoate)
• $[\text{HA}]$ is the concentration of the weak acid (benzoic acid)
• $K_a$ is the acid dissociation constant

Step 2: Calculate $pK_a$

$pK_a = -\log(K_a) = -\log(6.5 \times 10^{-5}) \approx 4.187$

Step 3: Plug in Values

$\text{pH} = 4.187 + \log\left(\frac{0.10}{0.15}\right)$

Calculating the logarithm:

$\log\left(\frac{0.10}{0.15}\right) = \log(0.6667) \approx -0.176$

Now, substitute:

$\text{pH} \approx 4.187 - 0.176 \approx 4.011$

Step 4: Calculate $[H^+]$

Using the pH to find the concentration of $H^+$:

$[H^+] = 10^{-\text{pH}} = 10^{-4.011} \approx 9.77 \times 10^{-5} \text{ M}$

Step 5: Calculate Ionization of Benzoic Acid

The percent ionization is calculated as:

[ \text{Percent Ionization} = \left(\frac{[H^+]}{[\text{HA}]_0}\right) \times 100% ]

Where $[\text{HA}]_0$ is the initial concentration of benzoic acid (0.15 M):

$\text{Percent Ionization} = \left(\frac{9.77 \times 10^{-5}}{0.15}\right) \times 100\% \approx 0.0651\%$

Conclusion

The percent ionization of the 0.15 M benzoic acid solution in a 0.10 M sodium benzoate solution is approximately 0.0651%.

Do you want further details or explanations? Here are some related questions:

1. What is the role of sodium benzoate in this solution?
2. How does pH affect the ionization of weak acids?
3. What assumptions are made in this calculation?
4. How would the percent ionization change with a different concentration of sodium benzoate?
5. Can you explain the concept of the Henderson-Hasselbalch equation further?

Tip: Always check the assumptions of your calculations, especially when dealing with equilibria in solutions.