Question

Aspirin is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin is prepared by dissolving 3.60 g/L. The pH of this solution is found to be 2.6. Calculate Ka for aspirin. (atomic mass: C

Answer 1

Answer:

Ka = 3.50x10⁻⁴

Explanation:

First, we need to convert the unit of 3.60 g/L to mol/L:

$C_{C_{9}H_{8}O_{4}} = 3.60 \frac{g}{L}*\frac{1 mol}{180.16 g} = 0.0200 mol/L$

The reaction dissociation of aspirin in water is:

C₉H₈O₄  +  H₂O  ⇄  C₉H₇O₄⁻ + H₃O⁺    

0.02 - x                        x             x

The constant of the above reaction is:

$Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]}$

$Ka = \frac{x^{2}}{0.02 - x}$

To find Ka we need to find the value of x. We know that pH = 2.6 so:

$pH = -log[H_{3}O^{+}]$

$2.6 = -log(x)$

$x = 2.51 \cdot 10^{-3} M = [H_{3}O^{+}] = [C_{9}H_{7}O_{4}^{-}]$

Now, the concentration of C₉H₈O₄ is:

$C_{C_{9}H_{8}O_{4}} = 0.02 - 2.51 \cdot 10^{-3} = 0.018 M$

Finally, Ka is:

$Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]} = \frac{(2.51 \cdot 10^{-3})^{2}}{0.018} = 3.50 \cdot 10^{-4}$

Therefore, the Ka of aspirin is 3.50x10⁻⁴.

       

I hope it helps you!