Chapter 17, Problem 50A

Interpretation Introduction

Interpretation:

The solubility of a given compound in gram per liter is to be calculated.

Concept Introduction :

The solubility product is the product of the concentration of its constituent ions with each ion concentration term raised to the power equal to the number of ions produced or dissociated.

Answer to Problem 50A

The solubility of ${\text{CaCO}}_{\text{3}}$ in g/L is $\text{0}{\text{.55×10}}^{\text{-3}}\text{g/L}$ .

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

  ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{CO}}_{\text{3}}{}^{\text{-2}}\left(\text{aq}\right)$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in an amount from an initial stateto equilibrium)

E $\text{= x x }$ (equilibrium concentration)

Given information:

Solubility product $\text{=3}{\text{.0×10}}^{\text{-9}}$

The Molar mass of calcium carbonate= $\text{100}\text{.0869}\text{g/mol}$

The solubility of the ions iscalculated as,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Ca}}^{\text{+2}}\right]\text{×}\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]\\ {\text{K}}_{\text{sp}}\text{=}\left(\text{x}\right)\text{×}\left(\text{x}\right)\\ {\text{K}}_{\text{sp}}\text{=}{\text{x}}^{\text{2}}\text{or}\\ \text{x}\text{=}\sqrt{{\text{K}}_{\text{sp}}}\\ \text{x}\text{=}\sqrt{{\text{3×10}}^{\text{-9}}}\\ \text{x}\text{=}\sqrt{{\text{30×10}}^{\text{-10}}}\\ \text{x}\text{=}\text{5}{\text{.5×10}}^{\text{-5}}\end{array}$

The solubility of calcium and carbonate ions in gram per liter is

  $\begin{array}{l}\text{S}\text{=}\text{5}{\text{.5×10}}^{\text{-5}}\\ \text{=}\text{5}{\text{.5×10}}^{\text{-5}}\text{×100}\text{.0869}\text{g/L}\\ \text{=}\text{0}{\text{.55×10}}^{\text{-3}}\text{g/L}\end{array}$

Conclusion

The solubility of calcium and carbonate ions is $\text{0}{\text{.55×10}}^{\text{-3}}\text{g/L}$ .