Question

Calculate the solubility of BaCO3 (a) in pure water and (b) in a solution in which [CO32-] = 0.289 M. Solubility in pure water = M Solubility in 0.289 M CO32- = M

Answer 1

(a). The solubility of
`BaCO_(3)` in pure water is
`4.4*10^(-5)\ M`

(b). The solubility of
`BaCO_(3)` in solution is
`6.92*10^(-9)\ M`

Step-by-step explanation:

Given that,

(a). The solubility of
`BaCO_(3)` in pure water

(b). The solubility of
`BaCO_(3)` in a solution

Solubility of
`CO_(3)^(-2)` is 0.289 M

We know that,

The solubility product constant of
`BaCO_(3)` is
`2*10^(-9)`

Let the solubility of
`BaCO_(3)` is s.

We need to calculate the solubility of
`BaCO_(3)` in pure water

Using formula of solubility

`ksp=s* s`

`ksp=s^2`

`s=√(ksp)`

Put the value into the formula

`s=\sqrt{2*10^(-9)}`

`s=4.4*10^(-5)\ M`

(b). We need to calculate the solubility of
`BaCO_(3)` in solution

Using formula of solubility

`ksp=s* s`

Put the value into the formula

`2*10^(-9)=s* 0.289`

`s=(2*10^(-9))/(0.289)`

`s=6.92*10^(-9)\ M`

Hence, (a). The solubility of
`BaCO_(3)` in pure water is
`4.4*10^(-5)\ M`

(b). The solubility of
`BaCO_(3)` in solution is
`6.92*10^(-9)\ M`