An unknown compound, X, is though to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 mL of 0.1M NaOH is added to 100 mL of a 0.1M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group of X. Can someone please tell me how do I know that at the initial pH the carboxylic acid will be 50% dissociated?
The Henderson-Hasselbalch equation is
pH = pKa + log(B/A)
pH = 2
pKa = 2
2 = 2 + log(B/A)
so B/A = 1 or base = acid
So RCOOH ==> RCOO^- + H^+
RCOO^- is base.
RCOOH is acid.
If base = acid it must be 50% dissociated.
To determine if the carboxylic acid will be 50% dissociated at the initial pH, we need to calculate the ratio of the concentrations of the dissociated species (A-) to the undissociated species (HA) at that pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given that the pKa of the carboxyl group is 2.0, we can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the values into the equation, we have:
[A-]/[HA] = 10^(2.0 - 2.0) = 10^0 = 1
So, at the initial pH of 2.0, the ratio [A-]/[HA] is 1, indicating that the carboxylic acid is 50% dissociated.