For this problem, we’re being asked to calculate the **solubility (in g/L) of AgCl(s)** in **0.0010 M MgCl_{2}(aq)**. Since the compounds are ionic compounds, they form ions when dissociating in water. The dissociation of

The chloride ion, Cl^{–}, has a charge of –1. Ag has a charge is +1:

AgCl(s) ⇌ Ag^{+}(aq) + **Cl ^{–}(aq)**

The chloride ion, Cl^{–}, has a charge of –1. Mg has a charge is +2:

MgCl_{2}(s) ⇌ Mg^{2+}(aq) + 2 **Cl ^{–}(aq)**

Calculate the molarity of Cl^{-}:

$\frac{\mathbf{0}\mathbf{.}\mathbf{0010}\overline{)\mathbf{}{\mathbf{MgCl}}_{\mathbf{2}}\mathbf{}\mathbf{mol}}}{\mathbf{1}\mathbf{}\mathbf{L}}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{Cl}}^{\mathbf{-}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{}{\mathbf{MgCl}}_{\mathbf{2}}\mathbf{}\mathbf{mol}}}$

**= 0.0020 M Cl**^{-}

Notice that there is a common ion present, **Cl ^{–}**. The

The K_{sp} of AgCl is 1.6 x 10^{-10}. What is the solubility of AgCl in 0.0010 M MgCl_{2}? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).

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