A buffer solution is prepared by dissolving 0.200 mol of NaH₂PO₄ and 0.100 mol of NaOH in enough water to make 1.00 L of solution.What is the pH of the H₂PO₄^-/HPO₄^2- buffer if the K_a2 = 6.2 × 10^-8? A)6.91 B)7.21 C)7.51 D)7.71

Question

Quizplus · Accepted Answer

First, we need to identify which species of the buffer is the acid and which is the base. In this case, NaHS is the acid and NaOH is the base.

Next, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

pH = pKa + log([base]/[acid])

where pKa is the acid dissociation constant, [base] is the concentration of the base (NaOH), and [acid] is the concentration of the acid (NaHS).

The pKa can be calculated from the given Ka (acid dissociation constant) value:

Ka = [H+][HS-]/[H2S]
pKa = -log(Ka) = 7.21

Now we plug in the values and solve for pH:

pH = 7.21 + log(0.100/0.200)
pH = 7.21 - 0.301 = 6.91

Therefore, the answer is B (6.91).

A Buffer Solution Is Prepared by Dissolving 0