I've read in chemwiki about the diverse ion effect. Assuming I have an aqueous solution of $\ce{NaCl}$ and an associated $K_\mathrm{sp}$. If I add $\ce{KI}$ there is an associated $K_\mathrm{sp}'$ for $\ce{NaCl}$ where $K_\mathrm{sp}' > K_\mathrm{sp}$. This actually means that I can dissolve more $\ce{NaCl}$ than I could before. Why does this happen? Shoudln't the new solution actually be closer to saturation and $K_\mathrm{sp}' < K_\mathrm{sp}$?
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To illustrate the "diverse ion effect", we will study how the solubility of $\ce{AgCl}$ changes in absence and presence of $\ce{NaNO3}$, given that the molar solubility product of $\ce{AgCl}$ is $1.76 \times 10^{-10}$.
The exact definition of the solubility constant is: $$K_{sp}= a(\ce{Ag+}).a(\ce{Cl-})=[\ce{Ag+}].\gamma(\ce{Ag+}).[\ce{Cl-}].\gamma(\ce{Cl-})$$ Where $a(\ce{Ag+})$ and $a(\ce{Cl-})$ are the activities of these ions, and $\gamma(\ce{Ag+})$ and $\gamma(\ce{Cl-})$
- In dilute solutions, i.e. absence of $\ce{NaNO3}$: $$\gamma(\ce{Ag+})=\gamma(\ce{Cl-})=1$$ So, the solubility of $\ce{AgCl}$ is equal to $$[\ce{Ag+}]= \sqrt{K_{sp}}= 1.33 \times 10^{-5}$$
- In presence of $\ce{NaNO3}$ of a concentration of $0.1 \mathrm{mol/kg}$, then $$[\ce{Ag+}]= \sqrt{K_{sp}}/ \sqrt{(\gamma(\ce{Na+}).\gamma(\ce{NO3-})} = 1.33 \times 10^{-5}/ (0.769)= 1.73 \times 10^{-5} $$ i.e. in the presence of a soluble salt, the molar solubility of a slightly soluble salt increases.
answered Sep 6 '15 at 19:26
