Chapter 17.3, Problem 21PP

Interpretation Introduction

Interpretation:

The solubility of lead carbonate is to be determined when solubility product is given at a particular temperature.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

The solubility product is given as

${\text{K}}_{\text{sp}}\text{= }\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]$

Answer to Problem 21PP

The solubility of lead carbonate in g/L is $\text{726}{\text{.8112 × 10}}^{\text{-4}}$ g/L.

Explanation of Solution

Given information : The solubility product of lead carbonate is $\text{7}{\text{.40×10}}^{\text{-4}}$.

$\text{PbCO3}$ Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

${\text{PbCO}}_{\text{3}}\rightleftharpoons {\text{Pb}}^{\text{2+}}{\text{+CO}}_{\text{3}}{}^{\text{-2}}$

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

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

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

The solubility product is given as

${\text{K}}_{\text{sp}}\text{= }\left[{\text{Pb}}^{\text{2+}}\right]\text{. }\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$

$\text{= x }\text{. x}$

${\text{= x}}^{\text{2}}$

$\sqrt{\text{Ksp}}\text{ = x}$

Or, $\text{x = }\sqrt{\text{7}\text{.40}}\times {\text{10}}^{\text{-14}}$

$\text{= 2}\text{.72}\times {\text{10}}^{}$ mol/L

Or, The solubility in terms of g/L is calculated as

= $\frac{\text{2}{\text{.72×10}}^{\text{-7}}\text{mol}}{\text{L}}\text{×}\frac{\text{267}\text{.21}\text{mol}}{\text{L}}$

$\text{= 726}{\text{.8112}}_{}{\text{x 10}}^{\text{-4}}$ g/L

Conclusion

The solubility of lead carbonate in g/L is $\text{726}{\text{.8112 × 10}}^{\text{-4}}$ g/L.