Question

question 3 of 10
what is $k_a$ for $h_3bo_3(aq)=h^+(aq)+h_2bo_3^-(aq)$?
a. $k_a = \frac{h_3bo_3}{h^+h_2bo_3^-}$
b. $k_a = \frac{h^+h_2bo_3^-}{h_3bo_3}$
c. $k_a = \frac{h^+h_3bo_3}{h_2bo_3^-}$
d. $k_a = h^+h_2bo_3^-$

Explanation:

Step1: Recall equilibrium - constant expression

For a general reaction $aA + bB
ightleftharpoons cC + dD$, the equilibrium - constant expression $K_{a}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$, where $[ ]$ represents the molar concentration of the species.
For the reaction $H_{2}BO_{3}^{-}(aq)
ightleftharpoons H^{+}(aq)+HBO_{3}^{2 - }(aq)$, the acid - dissociation constant $K_{a}$ is given by the ratio of the concentrations of the products to the concentration of the reactant.

Step2: Write the $K_{a}$ expression

The products are $H^{+}$ and $HBO_{3}^{2 - }$, and the reactant is $H_{2}BO_{3}^{-}$. So, $K_{a}=\frac{[H^{+}][HBO_{3}^{2 - }]}{[H_{2}BO_{3}^{-}]}$.

Answer:

B. $K_{a}=\frac{[H^{+}][HBO_{3}^{2 - }]}{[H_{2}BO_{3}^{-}]}$