Final answer:
By using the Henderson-Hasselbalch equation and given values of pH, pKa, and total concentration, we calculate the concentration of acetic acid to be 158.5 mM and the concentration of sodium acetate to be 91.5 mM in a 250 mM buffer solution with a pH of 5.0.
Step-by-step explanation:
To calculate the concentrations of acetic acid and sodium acetate in the buffer, we can use the Henderson-Hasselbalch equation:
\( pH = pKa + \log \frac{[A^-]}{[HA]} \)
Where \(pH\) is the pH of the buffer, \(pKa\) is the dissociation constant of acetic acid, \([A^-]\) is the concentration of the acetate ion (from sodium acetate), and \([HA]\) is the concentration of acetic acid.
We are given:
- \(pH = 5.0\)
- \(pKa = 4.75\)
- Total concentration of acid and conjugate base (\([HA] + [A^-]\)) = 250 mM
Firstly, substitute the pH and pKa into the equation and solve for the ratio \(\frac{[A^-]}{[HA]}\):
\(5.0 = 4.75 + \log \frac{[A^-]}{[HA]}\)
\(0.25 = \log \frac{[A^-]}{[HA]}\)
\(\frac{[A^-]}{[HA]} = 10^{0.25}\)
Now, let \(x\) represent the concentration of acetic acid (\(HA\)), and \((250 mM - x)\) represent the concentration of sodium acetate (\(A^-\)). We can set up the equation:
\(\frac{250 mM - x}{x} = 10^{0.25}\)
Upon solving for \(x\), we get:
\(x\) (concentration of acetic acid) = 158.5 mM
\((250 mM - x)\) (concentration of sodium acetate) = 91.5 mM