Question

What concentration of SO 2 − 3 SO32− is in equilibrium with Ag 2 SO 3 ( s ) Ag2SO3(s) and 7.10 × 10 − 3 7.10×10−3 M Ag + Ag+?

Answer 1

The question happens to be in an incorrect order but the correct question can be seen below;

What concentration of
`SO^(2-)_3` is in equilibrium with
`Ag_2SO_(3(S))` and
`7.10*10^(-3)M`
`Ag^+`? (The
`K_(sp)` of

Answer:

`2.96*10^(-10)M`

Step-by-step explanation:

The concentration of
`SO^(2-)_3` can be determined by using the solubility concept.

Given ionic solid is
`Ag_2SO_(3(S))` ;

The Equilibrium Equation for the ionic compound will be:

`Ag_2SO_(3(S))` ⇄
`2Ag_((aq))` +
`SO^(2-)_3_((aq))`

Now, the solubility product (
`K_(sp)`) of the ionic compound will be;

`K_(sp)`
`= [Ag^+]^(2)[SO^(2-)_3]`

Given that;

the concentration
`Ag^+` is
`7.10*10^(-3)M` ; &

solubility product of the given ionic solid is
`1.5*10^(-14)`

∴

`K_(sp)`
`= [Ag^+]^(2)[SO^(2-)_3]`

`1.5*10^(-14)`
`= (7.10*10^(-3))^2`
`[SO^(2-)_3]`

`[SO^(2-)_3]` =
`(1.5*10^(-14))/( (7.10*10^(-3))^2)`

= 2.97560008 × 10⁻¹⁰

≅
`2.96*10^(-10)M`

Thus, the concentration of
`[SO^(2-)_3]` is
`2.96*10^(-10)M`