Using Henderson-Hasselbalch equation, if pH = 5.00 and pKa = 4.74, what is the ratio [A−]/[HA]?

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Multiple Choice

Using Henderson-Hasselbalch equation, if pH = 5.00 and pKa = 4.74, what is the ratio [A−]/[HA]?

Explanation:
Henderson-Hasselbalch relates pH to the ratio of conjugate base to weak acid: pH = pKa + log([A−]/[HA]). Here, pH − pKa = 5.00 − 4.74 = 0.26, so log([A−]/[HA]) = 0.26. Exponentiating gives [A−]/[HA] = 10^0.26 ≈ 1.82. This means there is about 1.8 times as much conjugate base as its weak acid at this pH. If pH were equal to pKa, the ratio would be 1; if pH were lower, the ratio would be less than 1; if higher, greater than 1. The value around 1.82 corresponds to a pH a bit above the pKa, matching the given numbers.

Henderson-Hasselbalch relates pH to the ratio of conjugate base to weak acid: pH = pKa + log([A−]/[HA]). Here, pH − pKa = 5.00 − 4.74 = 0.26, so log([A−]/[HA]) = 0.26. Exponentiating gives [A−]/[HA] = 10^0.26 ≈ 1.82. This means there is about 1.8 times as much conjugate base as its weak acid at this pH. If pH were equal to pKa, the ratio would be 1; if pH were lower, the ratio would be less than 1; if higher, greater than 1. The value around 1.82 corresponds to a pH a bit above the pKa, matching the given numbers.

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