An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

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An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

d 8. When 75 mL of 0.1 M NaOH was added to 100mL of a 0.1 M solution of X at pH 2.0, the pH increased to 6.72. Calculate the pKa of the second group of X.
Source https://www.physicsforums.com/threads/calculating-pka.89490/

Chemistry

1 answer:

alisha [2.9K] · Answer 1 · 2026-02-21T19:28:55+03:00

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, $pH = -log[H^{+}]$ , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

$pH = pKa - log \frac{[HA]}{[A^{-}]}$

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of $OH^-$ become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of $OH^-$ reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of $OH^-$ , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

$6.72 = pKa - log \frac{0.01}{0.0025}$

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3