The pH of a 0.118 M solution of an aqueous weak acid (HA)is 3.20.The K_a for the weak acid is: A)5.4 $\times$ 10^-³ B)4.0 $\times$ 10^-⁷ C)3.4 $\times$ 10^-⁶ D)3.2 E)none of these

Question

Quizplus · Accepted Answer

Using the Henderson-Hasselbalch equation, we can find the pKa of the weak acid:
pH = pKa + log([A-]/[HA])
3.20 = pKa + log([A-]/0.118M)
Solving for pKa:
pKa = 3.4

Using the relationship between pKa and Ka:
pKa + pKb = 14
pKb = 10^(14-3.4) = 2.51 x 10^-11

Using the relationship between Ka and Kb for conjugate acid-base pairs:
Ka x Kb = Kw
Ka x 2.51 x 10^-11 = 1.0 x 10^-14
Ka = 3.98 x 10^-4 or 4.0 x 10^-4

Therefore, the best choice is C: 3.4 x 10^-4 (which has been rounded to one significant figure).

The PH of a 0 ×\times× 10-3 B)4 ×\times× 10-7 C)3 ×\times× 10-6 D)3

The PH of a 0 $\times$ 10^-³
B)4 $\times$ 10^-⁷
C)3 $\times$ 10^-⁶
D)3