Based on the acid strength of H2SO3 and H2SeO3, will the equilibrium constant of H2SO3(aq)+HSeO3−(aq)⇌H2SeO3(aq)+HSO 3−(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

Kc> 1, because is a stronger acid than

Explanation:

We have to start analyzing the reaction:

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

pKa of = 1.81

pKa of = 2.62

When we have a smallest pKa value we will have more acidity, therefore is a stronger acid than . 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 is a stronger acid than

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

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

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

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

the answer is either b or c

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.