Question

The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what is the value of Kc for the reaction, Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq) A) 9.6 × 10‒12 B) 7.7 C) 1.1 × 1023 D) 1.3 × 10‒1 E) 9.4 × 10‒24

Answer 1

Value of
`K_(c)` for the given reaction is 7.7

Step-by-step explanation:

`Ag_(2)CO_(3)(s)\rightleftharpoons 2Ag^(+)(aq.)+CO_(3)^(2-)(aq.)`

`K_(sp)(Ag_(2)CO_(3))=[Ag^(+)]^(2)[CO_(3)^(2-)]`

`Ag_(2)CrO_(4)(s)\rightleftharpoons 2Ag^(+)(aq.)+CrO_(4)^(2-)(aq.)`

`K_(sp)(Ag_(2)CrO_(4))=[Ag^(+)]^(2)[CrO_(4)^(2-)]`

Where
`K_(sp)` represents solubility product

For the given reaction,
`K_(c)=([CO_(3)^(2-)])/([CrO_(4)^(2-)])` (concentration of pure solids remain constant during reaction. Hence their concentration is taken as 1 to exclude them from equilibrium constant expression)

So,
`K_(c)=([Ag^(+)]^(2)[CO_(3)^(2-)])/([Ag^(+)]^(2)[CrO_(4)^(2-)])`

or,
`K_(c)=(K_(sp)(Ag_(2)CO_(3)))/(K_(sp)(Ag_(2)CrO_(4)))=(8.5* 10^(-12))/(1.1* 10^(-12))=7.7`

Hence option (B) is correct