# Question 77b85

Jun 21, 2017

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

$\underbrace{\text{HSO"_4^"-")_color(red)("stronger acid") + underbrace("CO"_3^"2-")_color(blue)("stronger base") ⇌ underbrace("SO"_4^"2-")_color(blue)("weaker base") + underbrace("HCO"_3^"-")_color(red)("weaker acid}}$

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