Question

The solubility of silver carbonate is 0.032 m at 20°c. calculate its ksp.

Answer 1

Answer: The solubility product for the given salt is
`1.31* 10^(-4)`

Step-by-step explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is represented as
`K_(sp)`

Silver carbonate is an ionic compound formed by the combination of 2 silver ions and 1 carbonate ions.

The equilibrium reaction for the ionization of silver carbonate follows the equation:

`Ag_2CO_3(aq.)\rightleftharpoons 2Ag^(+)(aq.)+CO_3^(2-)(aq.)`

1 mole of silver carbonate produces 2 moles of silver ions and 1 mole of carbonate ion

The expression of
`K_(sp)` for above equation is:

`K_(sp)=[2Ag^+]^2[CO_3^(2-)]`

We are given:

`[Ag^+]=(2* 0.032)=0.064M`

`[CO_3^(2-)]=0.032M`

Putting values in above equation, we get:

`K_(sp)=(0.064)^2* 0.032=1.31* 10^(-4)`

Hence, the solubility product for the given salt is
`1.31* 10^(-4)`

Answer 2

Ag2Co3 -----> 2Ag+ + CO3^-2

Ksp= (Ag)^2) ( CO3^-2)

by use of moles ratio the concentration of Ag+ = 2 x0.032 , and that of CO3^_3 = 0.032m

Ksp is therefore= ( 2 x0.032)^2 x 0.032= 1.31 x10^-4