# How do you find the solubility of silver bromide in 0.01 M NaBr(aq)? The solubility product of silver bromide is 7.7 xx 10^-13.

Mar 15, 2017

When finding the solubility of a compound in a solution, it is always easier to find its solubility in pure water first.

Define solubility as $s$, so that with the 1:1 stoichiometry in the reaction

${\text{AgBr"(s) rightleftharpoons "Ag"^(+)(aq) + "Br}}^{-} \left(a q\right)$,

the solubility product constant is

${K}_{s p} = 7.7 \times {10}^{- 13} = \left[{\text{Ag"^(+)]["Br}}^{-}\right]$

$= s \cdot s = {s}^{2}$.

Thus, we have that the solubility in pure water is:

$s = \sqrt{{K}_{s p}} = 8.775 \times {10}^{- 7}$ $\text{M}$ ${\text{Br}}^{-}$ or ${\text{Ag}}^{+}$

In $\text{NaBr} \left(a q\right)$, what we have is a new initial concentration of ${\text{Br}}^{-}$, in which the concentration of ${\text{Br}}^{-}$ is 1:1 with that of $\text{NaBr}$, and induces a skewed equilibrium upon being introduced into the previous aqueous solution.

Define a new solubility $s '$ so that for the following ICE table we have:

${\text{AgBr"(s) rightleftharpoons "Ag"^(+)(aq) + "Br}}^{-} \left(a q\right)$,

$\text{I"" "--" "" ""0 M"" "" "" ""0.01 M}$
$\text{C"" "--" "+s'" "" "" } + s '$
$\text{E"" "--" "" "s'" "" "" ""(0.01 + s') M}$

Notice how the equilibrium previously had ${K}_{s p} = s \cdot s = {s}^{2}$. This one instead is:

${K}_{s p} = 7.7 \times {10}^{- 13} = s ' \left(0.01 + s '\right)$

We assume that $s '$ is small compared to $0.01$ as ${K}_{s p} \text{<<} {10}^{- 5}$. Thus, we have a simplified expression:

$7.7 \times {10}^{- 13} \approx 0.01 s '$

$\implies \textcolor{b l u e}{s ' = 7.7 \times {10}^{- 11}}$ $\textcolor{b l u e}{\text{M}}$

The true answer would have been from solving

$0 = {\left(s '\right)}^{2} + 0.01 s ' - 7.7 \times {10}^{- 13}$

which would give an identical answer because the percent dissociation is much less than 5%.

Therefore, the introduction of $\text{NaBr}$ into solution decreased the solubility of ${\text{Br}}^{-}$ ions by a factor of $\approx 11396$. Does that fall in line with Le Chatelier's principle?