Answer:
Ka = 3.50x10⁻⁴
Explanation:
First, we need to convert the unit of 3.60 g/L to mol/L:
[tex] C_{C_{9}H_{8}O_{4}} = 3.60 \frac{g}{L}*\frac{1 mol}{180.16 g} = 0.0200 mol/L [/tex]
The reaction dissociation of aspirin in water is:
C₉H₈O₄ + H₂O ⇄ C₉H₇O₄⁻ + H₃O⁺
0.02 - x x x
The constant of the above reaction is:
[tex] Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]} [/tex]
[tex] Ka = \frac{x^{2}}{0.02 - x} [/tex]
To find Ka we need to find the value of x. We know that pH = 2.6 so:
[tex] pH = -log[H_{3}O^{+}] [/tex]
[tex] 2.6 = -log(x) [/tex]
[tex] x = 2.51 \cdot 10^{-3} M = [H_{3}O^{+}] = [C_{9}H_{7}O_{4}^{-}] [/tex]
Now, the concentration of C₉H₈O₄ is:
[tex] C_{C_{9}H_{8}O_{4}} = 0.02 - 2.51 \cdot 10^{-3} = 0.018 M [/tex]
Finally, Ka is:
[tex] 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} [/tex]
Therefore, the Ka of aspirin is 3.50x10⁻⁴.
I hope it helps you!
