Certainly! Let's solve the problem step-by-step.
Given:
- The concentration of the weak acid, [tex]\([HA]\)[/tex], is 0.210 M.
- The concentration of the conjugate base, [tex]\([A^-]\)[/tex], is 0.480 M.
- The acid dissociation constant, [tex]\(K_a\)[/tex], is [tex]\(3.0 \times 10^{-5}\)[/tex].
To find the pH of the buffer solution, we can use the Henderson-Hasselbalch equation:
[tex]\[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \][/tex]
Step 1: Calculate [tex]\(\text{p}K_a\)[/tex]
[tex]\[
\text{p}K_a = -\log_{10}(K_a)
\][/tex]
Given [tex]\(K_a = 3.0 \times 10^{-5}\)[/tex]:
[tex]\[
\text{p}K_a = -\log_{10}(3.0 \times 10^{-5}) \approx 4.522878745280337
\][/tex]
Step 2: Calculate the ratio of the conjugate base to the weak acid
[tex]\[
\frac{[A^-]}{[HA]} = \frac{0.480}{0.210} \approx 2.2857142857142856
\][/tex]
Step 3: Calculate the logarithm of the ratio
[tex]\[
\log_{10} \left( \frac{[A^-]}{[HA]} \right) = \log_{10}(2.2857142857142856) \approx 0.3590219426416697
\][/tex]
Step 4: Apply the Henderson-Hasselbalch equation
[tex]\[
\text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right)
\][/tex]
[tex]\[
\text{pH} = 4.522878745280337 + 0.3590219426416697 \approx 4.881900687922007
\][/tex]
So, the pH of the buffer solution is approximately [tex]\(4.8819\)[/tex].
