# What volume of water is necessary to dissolve 0.010 mole of AgCl?

## ${K}_{s P} i s$1.7*!0^-12#.

Dec 19, 2016

Around $77$ ${m}^{3}$; an impossibly large volume.

#### Explanation:

We have the ${K}_{\text{sp}}$, and clearly we must convert this to a solubility in $g \cdot {L}^{-} 1$.

$A g C l \left(s\right) r i g h t \le f t h a r p \infty n s A {g}^{+} + C {l}^{-}$

${K}_{s p} = \left[A {g}^{+}\right] \left[C {l}^{-}\right] = 1.7 \times {10}^{-} 12$

If we call $S$ the solubility of $A g C l$, then $S = \left[A {g}^{+}\right] = \left[C {l}^{-}\right]$, and $S = \sqrt{{K}_{s p}} = S = \sqrt{1.7 \times {10}^{-} 12} = 1.303 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$.

And then we convert this molar solubility into a solubility in $g \cdot {L}^{-} 1$:

$\text{Solubility}$ $=$ $1.303 \times {10}^{-} 6 \cdot \cancel{m o l} \cdot {L}^{-} 1 \times 143.32 \cdot g \cdot \cancel{m o {l}^{-} 1}$

$= 1.87 \times {10}^{-} 5 \cdot g \cdot {L}^{-} 1$. Again, this is a low value.

You required a solution of a molar quantity of $0.01 \cdot m o l$, i.e. $\frac{1.43 \cdot g}{1.87 \times {10}^{-} 5 \cdot g \cdot {L}^{-} 1} = 76.4 \times {10}^{3} \cdot L = 76.4 \cdot {m}^{3}$, which is a rather large volume, and the problem is completely impractical.

So what would a chemist do if he or she wanted to get silver ion into solution from silver halide? They would probably use a complexing ligand, for instance, ${S}_{2} {O}_{3}^{2 -}$ or $N {H}_{3}$ or something that would allow the silver ion to dissolve in solution in the presence of the halide that would normally precipitate it out.