Question

What is the pKa of an acid if a buffer made from 0.0960 mol of the acid and 0.0450 mol of its conjugate base in water has a pH of 3.75?

(a) 3.46
(b) 3.62
(c) 4.00
(d) 3.75
(e) 4.08

Answer 1

The pKa of an acid used to make a buffer solution with a pH of 3.75, having 0.0960 mol of the acid and 0.0450 mol of its conjugate base, is calculated using the Henderson-Hasselbalch equation to be 4.08.

Step-by-step explanation:

The question asks to determine the pKa of an acid used to make a buffer solution with given concentrations of the acid and its conjugate base. The pH of the buffer solution is 3.75. By using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid, we can solve for the pKa. Since the pH and the ratio of the base to acid concentrations are given, the equation can be rearranged to find the pKa:

pKa = pH - log([A-]/[HA])

pKa = 3.75 - log(0.0450/0.0960)

pKa = 3.75 - log(0.46875)

pKa = 3.75 - (-0.329) = 4.08

Therefore, the pKa of the acid is 4.08, which corresponds to option (e).