Question

The half equivalence point in the titration of 0.100 m hco2h (ka = 1.8 × 10-4) with 0.250 m naoh occurs at ph = ________.

A) 3.33
B) 3
C) 3.9
D) 3.75

Answer 1

The half equivalence point occurs when pH equals the pKa of the acid. For HCO2H with Ka = 1.8 × 10^-4, the pKa is calculated as 3.74, making the closest answer D) 3.75.

Step-by-step explanation:

The student is asking about the half equivalence point of a titration process involving a weak acid, HCO2H (formic acid), with a known Ka value, by a strong base NaOH. The half equivalence point in a titration is the point where half of the acid has been neutralized by the base.

At the half equivalence point, the concentration of the acid equals the concentration of its conjugate base, and thus, the pH equals the pKa of the weak acid. The pKa can be calculated using the formula pKa = -log(Ka). Plugging in the given Ka value for formic acid (1.8 × 10-4), we find that pKa = -log(1.8 × 10-4) = 3.74. Therefore, the answer is D) 3.75, as it is the closest pH value to the calculated pKa.