Silver chloride (AgCI) is relatively insoluble in water. At 25 C, 1.3 x 10^3 L of water is needed to dissolve 2.5 g of AgCl. What mass (in milligrams) of AgCl will dissolve in 1.0 L of water?

Sep 5, 2016

Approx. $2 \cdot m g$

Explanation:

$\text{Solubility}$ $=$ $\text{Moles of solute"/"Volume of solution}$.

For silver chloride we are given that: $2.5 \cdot g$ wil dissolve in over $1$ ${m}^{3}$ of water.

And thus, $\text{Solubility}$ $=$ $\frac{2.5 \cdot g}{143.32 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1.3 \times {10}^{3} \cdot L}$ $=$ $1.34 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

And thus silver chloride solute in 1L of saturated water $=$ $1.34 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1 \times 143.32 \cdot g \cdot m o {l}^{-} 1 \times 1 L$ $\cong$ $2 \cdot m g$

Typically, you are quoted a ${K}_{\text{sp}}$ for these insoluble salts. ${K}_{\text{sp }} A g C l = 1.77 \times {10}^{-} 10$. ${K}_{\text{sp}}$ values can be shown to depend on the molar solubilities.