Solubility

Tutorial 11

    

A sodium chloride solution containing the chloride ion Cl^- is added dropwise to a solution containing the ion $Ag^+$. As long as the product $[Ag^+][Cl^-]$ has a lower value than the solubility product, no .......... is formed. However, once the product $[Ag^+][Cl^-]$ has attained the value of the solubility product, any attempt to increase the concentration of $[Ag^+][Cl^-]$ or $[Ag^+][Cl^-]$ will lead to the precipitation of AgCl. When $Cl^-$ or $Ag^+$ is added, the product $[Ag^+][Cl^-]$ remains .........., but the quantity of $AgCl(s)$ ........... Let $x$ be the maximum molarity of $NaCl$ in an $10^{-4}$ M solution of Ag⁺. Knowing that the solubility product of $AgCl$ is equal to $K_s$ $=$ $1.6\cdot10^{-10}$, calculate $x$.

Calculation: ....... We have $x$ $=$ $[Cl^-]$, because the chloride ions come from $NaCl$ alone. ....... In the saturated solution: $K_s$ $=$ $[Ag^+][Cl^-]$ $1.6\cdot 10^{-10}$ $=$ $10^{-4}x$ ....... Maximum molarity of $NaCl\;$ = $\frac{1.6\cdot 10^{-10}}{10^{-4}}$ $=$ $1.6\cdot 10^{-6}\frac{mol}{L}$

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