Question

Based on the acid strength of H2SO3 and H2SeO3, will the equilibrium constant of H2SO3(aq)+HSeO3−(aq)⇌H2SeO3(aq)+HSO3−(aq) be (Kc> 1) or (Kc< 1)? Group of answer choices Kc< 1, because H2SO3 is a stronger acid than H2SeO3 Kc< 1, because H2SO3 is a weaker acid than H2SeO3 Kc> 1, because H2SO3 is a weaker acid than H2SeO3 Kc> 1, because H2SO3 is a stronger acid than H2SeO3

Answer 1

Answer:

Kc> 1, because $H_2SO_3$ is a stronger acid than $H_2SeO_3$

Explanation:

We have to start analyzing the reaction:

$HSeO_3^-~+~H_2SO_3~~H_2SeO_3~+~HSO_3^-$

When we search the pKa values for each we will find:

pKa of $H_2SO_3$= 1.81

pKa of $H_2SeO_3$= 2.62

When we have a smallest pKa value we will have more acidity, therefore $H_2SO_3$ is a stronger acid than $H2SeO3$. Additionally, if we have his behavior the reaction would be displaced to the right side due to the strong acid (the strong acid is placed in the reactives side), so we will have a K greater than 1.  

The answers are:

a) Kc< 1, because $H_2SO_3$ is a stronger acid than $H_2SeO_3$

b) Kc< 1, because $H_2SO_3$ is a weaker acid than $H_2SeO_3$

c) Kc> 1, because $H_2SO_3$ is a weaker acid than $H_2SeO_3$

d) Kc> 1, because $H_2SO_3$ is a stronger acid than $H_2SeO_3$

The answer that fits with the previous analysis is "d". The molecule $H_2SO_3$  is the strongest acid and K is greater than one