Question

Question #77b85

Answer 1

The position of equilibrium lies primarily to the right.

Explanation:

`"HSO"_4^"-"color(white)(l) (K_text(a) = 1.2 × 10^"-2")` is a stronger acid than `"HCO"_3^"-"color(white)(l) (K_text(a) = 4.8 × 10^"-11")`.

Also,

`"CO"_3^"2-"color(white)(l) (K_text(b) = 2.1 × 10^"-4")` is a stronger base than `"SO"_4^"2-"color(white)(l) (K_text(b) = 8.3 × 10^"-13")`.

Thus, we have

`underbrace("HSO"_4^"-")_color(red)("stronger acid") + underbrace("CO"_3^"2-")_color(blue)("stronger base") ⇌ underbrace("SO"_4^"2-")_color(blue)("weaker base") + underbrace("HCO"_3^"-")_color(red)("weaker acid")`

Since the reactants are both stronger than the products, the position of equilibrium lies to the right.