To determine the pH of the buffer solution created by mixing equal volumes of a 2.0 M HC2H3O2 (acetic acid) and 0.5 M NaC2H3O2 (sodium acetate), we will use the Henderson-Hasselbalch equation. The equation is:
[tex]\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \][/tex]
- Here, [tex]\(\text{pKa}\)[/tex] is the acid dissociation constant of the weak acid.
- [tex]\([\text{A}^-]\)[/tex] is the concentration of the conjugate base (acetate ion, [tex]\(C_2H_3O_2^-\)[/tex]).
- [tex]\([\text{HA}]\)[/tex] is the concentration of the weak acid (acetic acid).
We are given:
- The [tex]\(\text{pKa}\)[/tex] of acetic acid is approximately 4.75.
- The concentration of acetic acid before mixing is 2.0 M.
- The concentration of sodium acetate before mixing is 0.5 M.
- Equal volumes of acetic acid and sodium acetate are mixed.
When we mix equal volumes, the concentration of each will be halved due to dilution since the volume doubles but the amount of substance stays the same. Therefore, the concentration of the acetic acid after mixing will be half of 2.0 M, which is 1.0 M. Similarly, the concentration of the sodium acetate after mixing will be half of 0.5 M, which is 0.25 M.
Plugging these values into the Henderson-Hasselbalch equation, we have:
[tex]\[ \text{pH} = 4.75 + \log \left(\frac{0.25}{1.00}\right) \][/tex]
We can simplify the logarithmic term:
[tex]\[ \log \left(\frac{0.25}{1.00}\right) = \log(0.25) \][/tex]
The logarithm of 0.25 to the base 10 is approximately -0.6 (as [tex]\(10^{-0.6} \approx 0.25\)[/tex]). Now, we plug this into the equation:
[tex]\[ \text{pH} = 4.75 - 0.6 \][/tex]
[tex]\[ \text{pH} = 4.15 \][/tex]
So, the pH of the buffer solution is approximately 4.15.
