Final answer:
The pH of the solution containing equal concentrations of acetic acid and its conjugate base is 4.74, using the Henderson-Hasselbalch equation. After adding 1.0 mL of 0.10 M NaOH, you would calculate the new acid and base concentrations to determine the new pH of the buffer solution.
Step-by-step explanation:
The pH of a buffer solution containing 0.10 M HC2H3O2 (acetic acid) and 0.10 M NaC2H3O2 (sodium acetate, the conjugate base of acetic acid) can be found using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm of the Ka value of acetic acid, [A-] is the concentration of the acetate ion, and [HA] is the concentration of acetic acid.
Given the Ka of acetic acid is 1.8 x 10-5, we calculate pKa as follows:
pKa = -log(1.8 x 10-5) = 4.74
Since the concentrations of the acetic acid and its conjugate sodium acetate are equal (0.10 M each), the log term becomes log(1) which is 0. Therefore, the pH of the solution is equal to the pKa of acetic acid, which is 4.74.
For the second part, after adding 1.0 mL of 0.10 M NaOH to 100 mL of this buffer, we need to compute the new concentrations:
moles of NaOH added = 1.0 mL x 0.10 M / 1000 = 1.0 x 10-4 moles.
Since NaOH is a strong base, it will react completely with acetic acid to form the conjugate acetate ion and water:
HC2H3O2 + OH- → C2H3O2- + H2O
This will decrease the acetic acid concentration by 1.0 x 10-4 moles and increase the acetate concentration by the same amount. After doing the appropriate calculations for the new concentrations, we would use the Henderson-Hasselbalch equation again to find the new pH.