Question

What is the solubility of MgCO₃ in a solution that contains 0.080 M Mg²⁺ ions? (Ksp of MgCO₃ is 3.5 × 10⁻⁸)

Answer 1

Answer: 4.4 x 10^-7

Explanation:

The dissociation equation for this reaction is:

MgCO3 (s) → Mg+2 (aq) + CO3-2 (aq)

$\text { So, } k_{s p}=(x+0.08) x$ (Here 0.08 >>> x )

$\begin{aligned}\Rightarrow 3.5 \times 10^{-8} &=0.08 \times x \\\Rightarrow x &=\frac{3.5}{0.08} \times 10^{-8} \\&=4.37 \times 10^{-7} \\\Rightarrow x & \approx 4.4 \times 10^{-7} \mathrm{~M}\end{aligned}$

So the solubility MgCO₃ in a solution that containing 0.080 M Mg²⁺ is 4.4 x 10^-7