How to find ion concentrations at various points in a solution that will precipitate multiple insoluble salts?


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> * The solubility product constant of lead (II) iodide is $1.2 \times 10^{-5}$ at 273 K.
> * The solubility product constant for lead (II) carbonate is $1.5 \times 10^{-13}$ at the same temperature.
> * 50 milliliters of a solution at this temperature contains an unknown concentration of carbonate ions and 0.050 M iodine ion concentration.
> * A secondary solution of 0.10 M lead (II) nitrate is added. After 32 milliliters of the lead (II) nitrate solution is added, the lead (II) iodide begins to precipitate.
> * Calculate the original concentration of the carbonate ions in addition to the concentration of carbonate ions in the solution when the lead (II) iodide begins to precipitate.
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I attempted to find the concentration of carbonate ions when the lead iodide started to precipitate by dividing the lead carbonate's solubility product constant by the concentration of lead ions at this time, since the lead carbonate precipitates first.


So I concluded that it will be at equilibrium at this point ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$ and then multiplying this by the current volume (0.082 L) and dividing by the original volume (0.05 L) to get the original concentration of carbonate ions, at $6.3 \times 10^{-12}$ M.


To get the concentration at the time when the lead iodide started dissolving, I took my answer from the first part of the earlier calculation ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$, for a final carbonate ion concentration of $3.8 \times 10^{-12}$. However, I am unsure if the process I used was correct.


Any help is appreciated. Thank you.




Your reasoning is too much complicated. If all the data are calculated in moles, the result is immediate. Let's try !


$32$ mL of $0.1 M \ce{Pb^{2+}}$ contains $0.0032$ mol $\ce{Pb}$. As a consequence, the initial carbonate solution contains the same amount of $\ce{CO3^{2-}}$. As these $0.0032$ mole are dissolved in $50$ mL. It means that the initial concentration $\ce{[CO3^{2-}]}$ is $\ce{0.0032 mol/0.05 L = 0.064 M}$


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