A 0.10 M solution of a weak monoprotic acid has a hydronium-ion concentration of 8.0 × 10^-4 M.What is the acid-ionization constant,K_a,for this acid? A) 2.8 × 10^-2 B) 7.5 × 10^-3 C) 8.0 × 10^-4 D) 6.4 × 10^-6 E) 9.6 × 10^-5

Question

Quizplus · Accepted Answer

The acid-ionization constant, $K_{a}$, can be calculated using the formula:

$$K_{a}=\frac{[	ext{H}_{3}	ext{O}^{+}][	ext{A}^{-}]}{[	ext{HA}]}$$

where [HA] is the initial concentration of the weak acid, [A$^-$] is the concentration of the conjugate base, and [H$_{3}$O$^{+}$] is the hydronium ion concentration.

In this case, we have:

[HA] = 0.10 M (the concentration of the weak acid)
[H$_{3}$O$^{+}$] = 8.0 $	imes$ 10$^{-5}$ M (the hydronium ion concentration)

To find [A$^-$], we can use the fact that weak acids only partially dissociate in water and set [A$^-$] approximately equal to the concentration of H$_{3}$O$^{+}$:

$$[	ext{A}^{-}]\approx[	ext{H}_{3}	ext{O}^{+}]=8.0 	imes 10^{-5}$$

Substituting these values into the expression for $K_{a}$ and solving for the unknown gives:

$$K_{a}=\frac{(8.0 	imes 10^{-5})^{2}}{0.10-8.0 	imes 10^{-5}}=6.4 	imes 10^{-6}$$

Therefore, the answer is (D): $6.4 	imes 10^{-6}$.

A 010 M Solution of a Weak Monoprotic Acid Has a a Hydronium-Ion