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

Answer:

(a). The solubility of $BaCO_{3}$ in pure water is $4.4\times10^{-5}\ M$

(b). The solubility of $BaCO_{3}$ in solution is $6.92\times10^{-9}\ M$

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 $2\times10^{-9}$

Let the solubility of $BaCO_{3}$ is s.

We need to calculate the solubility of $BaCO_{3}$ in pure water

Using formula of solubility

$ksp=s\times s$

$ksp=s^2$

$s=\sqrt{ksp}$

Put the value into the formula

$s=\sqrt{2\times10^{-9}}$

$s=4.4\times10^{-5}\ M$

(b). We need to calculate the solubility of $BaCO_{3}$ in solution

Using formula of solubility

$ksp=s\times s$

Put the value into the formula

$2\times10^{-9}=s\times 0.289$

$s=\dfrac{2\times10^{-9}}{0.289}$

$s=6.92\times10^{-9}\ M$

Hence, (a). The solubility of $BaCO_{3}$ in pure water is $4.4\times10^{-5}\ M$

(b). The solubility of $BaCO_{3}$ in solution is $6.92\times10^{-9}\ M$