Question

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 and 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/

Answer 1

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that $pH = -log[H^{+}]$, and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

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

Where [HA] is the concentration of the acid, and $[A^{-}]$ is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of $OH^-$ will be

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

So, the 0.0075 mol of $OH^-$ reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of $OH^-$, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

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

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3