Question

A buffer was prepared with kh2po4 and k2hpo4 at a ph of 7.45. for h2po4⎯, the ka is 6.2 x 10⎯8. part a: what ratio of [a⎯]/[ha] was required to prepare this buffer?

Answer 1

To prepare a buffer at pH 7.45 with KH2PO4 and K2HPO4, when the Ka for H2PO4- is 6.2 x 10^-8, the required [A-]/[HA] ratio is approximately 1.74, derived using the Henderson-Hasselbalch equation.

Step-by-step explanation:

To determine the ratio of [A−]/[HA] required to prepare a buffer at a pH of 7.45 using KH2PO4 (acid, HA) and K2HPO4 (base, A−) when the Ka for H2PO4− is 6.2 x 10−8, we use the Henderson-Hasselbalch equation:

pH = pKa + log([A−]/[HA])

First, we calculate the pKa:

pKa = -log(Ka)

pKa = -log(6.2 x 10−8)

pKa = 7.21

Next, we substitute the pH and pKa into the Henderson-Hasselbalch equation and solve for the [A−]/[HA] ratio:

7.45 = 7.21 + log([A−]/[HA])

log([A−]/[HA]) = 7.45 - 7.21

log([A−]/[HA]) = 0.24

Therefore, [A−]/[HA] = 100.24

[A−]/[HA] ≈ 1.74

The required ratio of [A−]/[HA] to create a buffer solution at pH 7.45 with the specified components is approximately 1.74.