For a buffer with pKa = 4.75, what [A-]/[HA] ratio yields pH = 3.75?

For a buffer with pKa = 4.75, what [A-]/[HA] ratio yields pH = 3.75?

• A. 0.10
• B. 1.0
• C. 10
• D. 100

Answer: (A)

The Henderson–Hasselbalch relationship tells us how pH relates to pKa and the ratio of the conjugate base to the weak acid in a buffer. It states pH = pKa + log([A-]/[HA]). Plugging in the values, pH − pKa = 3.75 − 4.75 = −1.0, so log([A-]/[HA]) = −1. Therefore [A-]/[HA] = 10^(−1) = 0.1. This means there is ten times more A– (the base form) needed to be present for pH to rise to pKa, but at pH one unit below pKa, the acid form HA is more abundant, giving a ratio of 0.1.

The Henderson–Hasselbalch relationship tells us how pH relates to pKa and the ratio of the conjugate base to the weak acid in a buffer. It states pH = pKa + log([A-]/[HA]).

Plugging in the values, pH − pKa = 3.75 − 4.75 = −1.0, so log([A-]/[HA]) = −1. Therefore [A-]/[HA] = 10^(−1) = 0.1.

This means there is ten times more A– (the base form) needed to be present for pH to rise to pKa, but at pH one unit below pKa, the acid form HA is more abundant, giving a ratio of 0.1.