# Solutions of several salts

What happens when you have a solution of a salt, and then try to dissolve another salt in that solution? I know about the common ion-effect etc, but how do you solve problems like this:

How much AgCl can you solve per L in a 3.8g/L NaCl solution?

To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,

$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$

This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.

If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.

$$\begin{array}{lcl} x&=&[\ce{Ag+}]=[\ce{Cl-}]\\ x^2 &=& 1.8\times 10^{-10} \\ x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5} \end{array}$$

We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.

However, if we already have some $\ce{Cl-}$ in solution, then things are a little different. Let's say we have $3.8\ \mathrm{g/L} = 6.5\times10^{-2}\ \mathrm{mol/L}\ \ce{NaCl}$ in solution, now we have a different concentration of $\ce{Cl-}$. When $\ce{AgCl}$ is added, some will dissolve, and the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ will both increase by some amount $x$.

We solve these problems using the ICE (Initial, Change, Equilibrium} approach:

$$\begin{array}{c|r|l} \ & [\ce{Ag+}] &[\ce{Cl-}] \\ \hline \mathrm{I} & 0 & 6.5\times 10^{-2} \\ \mathrm{C} & +\ x & +\ x \\ \mathrm{E} & x & 6.5\times 10^{-2} + x\\ \hline \end{array} \\ \ \\ K_{\mathrm{sp}}=1.8\times 10^{-10}=x(6.5\times 10^{-2} + x)\\ 1.8\times 10^{-10}=x^2 +(6.5\times 10^{-2})x \\ 0 = x^2 +(6.5\times 10^{-2})x - 1.8\times 10^{-10}\\ x = 2.77\times 10^{-9}, -0.065$$

One of the values for $x$ is nonsensical. The concentration of $\ce{Ag+}$ cannot be negative. Thus the amount of $\ce{AgCl}$ that dissolved in one liter was $9.77\times 10^{-9}\ \mathrm{mol}=1.38\times 10^{-6}\ \mathrm{g}$.

Note that since $x\ll 6.5\times 10^{-2}$, we could have simplified to $6.5\times 10^{-2}+x \approx 6.5\times 10^{-2}$ and then our maths would not have needed to invoke the quadratic theorem while getting the same answer within significant figures. This approximation is often valid. However, unless you are taking a test, you should always be able to Wolfram Alpha.