Question

Consider the titration of a 23.0-ml sample of 0.105 m hc2h3O₃ (ka=1.8×10-5) with 0.120 m NaOH. Determine the pH at one-half of the equivalence point.

Answer 1

At one-half of the equivalence point in the titration of acetic acid with NaOH, the pH of the solution is equal to the pKa of acetic acid, which is 4.74.

Step-by-step explanation:

The student's question is related to the pH calculation at one-half of the equivalence point during the titration of a weak acid (acetic acid, HC2H3O2) with a strong base (NaOH). To determine the pH at the half-equivalence point, we can use the fact that the pH equals the pKa at this point in the titration. The given acid dissociation constant (Ka) for acetic acid is 1.8×10-5. Taking the negative logarithm to find pKa, we have:

pKa = -log(Ka) = -log(1.8×10-5)

Using the provided value of Ka, the calculation gives us:

pKa = 4.74

Thus, at one-half of the equivalence point, the pH of the solution will be 4.74.