# Given that K_"sp"=4xx10^-13 for "silver bromide", what will be the concentration of silver ion if [Br^-]=1xx10^-6*mol*L^-1?

May 5, 2017

We interrogate the solubility equilibrium.........

#### Explanation:

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

And thus $\left[A {g}^{+}\right] \left[B {r}^{-}\right] = {K}_{\text{sp}} = 4 \times {10}^{-} 13$.

And in pure solvent, $\left[A {g}^{+}\right] = \left[B {r}^{-}\right] = \sqrt{{K}_{\text{sp}}}$

But here $\left[B {r}^{-}\right]$ has been artificially raised so that the silver salt is LESS soluble:

$\left[B {r}^{-}\right] = \frac{4 \times {10}^{-} 13}{\left[A {g}^{+}\right]} = \frac{4 \times {10}^{-} 13}{1 \times {10}^{-} 6} = 4 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.

Such a phenomenon is known as $\text{salting out}$. If you had metals ions that were more valuable than silver, say those of rhodium, or iridium, or gold, you want to make sure that you had precipitated all the metal out. Likewise, if you had cadmium or mercury, you would also want to remove these salts from solution before disposal.