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 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group.

Answer 1

Answer:

pKa = 7.2

Explanation:

When pH=pKa=2.0, there are equal amounts of the X carboxyl group and its ionized form.

• 100 mL * 0.1 M = 10 mmol compound X

• 5 mmol ionized carboxyl X-COO⁻  & 5 mmol unionized carboxyl X-COOH

Then, (75 mL * 0.10 M) 7.5 mmol of OH⁻ are added, so all 5 mmol of X-COOH converts into X-COO⁻. Then the remaining (7.5 - 5) 2.5 mmol of OH⁻ react with the second ionizable group of X

Number of X-COO⁻ = 5mmol (from the beginning) + 5mmol (from X-COOH that reacted) - 2.5 mmol (from the OH⁻ remaining) = 7.5 mmol

Because the total number of X compound moles did not change, we have (10 - 7.5) 2.5 mmol of the conjugate base of X-COO⁻.

Now we have all required data to solve this problem using Henderson-Hasselbach's equation:

pH = pKa + log [A⁻]/[HA]

6.72 = pKa + log (2.5/7.5)

pKa = 7.2

Answer 2

Answer:

$pK_a$  = 7.20

Explanation:

Given that our Molarity of X(unknown compound)= 0.10 M

Volume of X = 100 ml = 0.1  L

Molarity  = $\frac{number of moles of X}{Volume of X}$

Number of moles of X = Molarity × Volume of X

= 0.1 M × 0.1 L

= 0.01 mol

$pK_a$ = $pH$

∴$[H^+]$ =$[HA]$    (this literaly implies and point out that the beginning the amount of carboxyl group and the second ionizable group must be equal.)

Having said that,

Number of moles of carboxyl group in X = $\frac{0.01mol}{2}$

= 0.005 mol

When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X;

we have the number of moles of NaOH that is being added as:

Molarity × volume of NaOH

= 0.1 M × 0.075 L

= 0.0075 mol

From the question, if NaOH molecules thoroughly dissociate the carboxyl group of X.

The excess NaOH can be calculated as:

Excess of (NaOH) = Number of moles of NaOH added - Number of moles of carboxyl group in X

Excess of (NaOH) = 0.0075 -0.005 = 0.0025 mol

∴ Applying Henderson-Hesselbalch equation; it will be easier to determine the $pK_a$ for second group:

$pK_a$  = $pH$ $-\frac{log[A]}{HA}$

       = 6.72 - log$(\frac{0.0025}{0.0075})$

       = 6.72 - log (0.333)

       = 6.72 + 0.477

$pK_a$  = 7.197

≅ 7.20