Final answer:
To find the concentrations of acetic acid and acetate ion in a buffer solution with given pH and pKa, we use the Henderson-Hasselbalch equation. After calculations, both HOAc and OAc- are found to be 0.1 M in the 0.2 M acetate buffer at pH 5.0.
Step-by-step explanation:
The student's question asks about the concentrations of acetic acid (HOAc) and acetate ion (OAc-) in a 0.2 M acetate buffer with a pH of 5.0, given that the pKa for acetic acid is 4.76. To calculate the concentrations of the acid and its conjugate base in a buffer solution, we use the Henderson-Hasselbalch equation:
\[ pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where \( pH \) is the pH of the buffer, \( pKa \) is the dissociation constant of the weak acid, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid. Rearranging the equation allows us to solve for \([A^-]\) and \([HA]\):
\[ [A^-] = [HA] \times 10^{(pH-pKa)} \]
Plugging in the given pH, pKa, and total concentration:
\[ [A^-] = [HA] \times 10^{(5.0-4.76)} \]
Since the total concentration \([A^-] + [HA]\) is 0.2 M, we can set up the following system of equations:
\[ [HA] + [A^-] = 0.2 M \]
\[ [A^-] = [HA] \times 10^{0.24} \]
Solving this system, we find the concentration of acetic acid (HOAc) and acetate ion (OAc-):
HOAc: 0.1 M
OAc-: 0.1 M