Using Henderson-Hasselbalch equation, which expression correctly relates pH, pKa, and the ratio of conjugate base to weak acid?

Using Henderson-Hasselbalch equation, which expression correctly relates pH, pKa, and the ratio of conjugate base to weak acid?

• A. pH = pKa - log([A-]/[HA])
• B. pH = pKa + log([HA]/[A-])
• C. pH = pKa + log([A-]/[HA])
• D. pH = pKa - log([HA]/[A-])

Answer: (C)

The Henderson-Hasselbalch relationship shows how pH depends on the acid’s dissociation constant and the ratio of conjugate base to weak acid. The correct form is pH = pKa + log([A-]/[HA]). Here, A- is the conjugate base and HA is the weak acid. The idea comes from Ka = [H+][A-]/[HA]; taking negative logs and rearranging gives pH = pKa + log([A-]/[HA]). This means that increasing the amount of conjugate base relative to the acid raises pH (the log term becomes larger), while equal amounts give pH = pKa, and more acid relative to base lowers pH (the log term is negative). The other forms would either subtract the log or invert the ratio, which would not correctly reflect how pH changes with the base/acid balance.

The Henderson-Hasselbalch relationship shows how pH depends on the acid’s dissociation constant and the ratio of conjugate base to weak acid. The correct form is pH = pKa + log([A-]/[HA]). Here, A- is the conjugate base and HA is the weak acid. The idea comes from Ka = [H+][A-]/[HA]; taking negative logs and rearranging gives pH = pKa + log([A-]/[HA]).

This means that increasing the amount of conjugate base relative to the acid raises pH (the log term becomes larger), while equal amounts give pH = pKa, and more acid relative to base lowers pH (the log term is negative). The other forms would either subtract the log or invert the ratio, which would not correctly reflect how pH changes with the base/acid balance.