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Question: How do you find the solubility of silver bromide in \( 0.01M \) \( NaBr(aq) \) ? The solubility prod...

How do you find the solubility of silver bromide in 0.01M0.01M NaBr(aq)NaBr(aq) ? The solubility product of silver bromide is 7.7×10137.7 \times {10^{13}} .

Explanation

Solution

Hint : To solve this question, firstly we will go through the stoichiometric ratio of the given reaction, and then we will find the new concentration of the component of the compound to find the solubility of the given silver bromide.

Complete Step By Step Answer:
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:
AgBr(s)⇌Ag+(aq)+Br−(aq)AgBr(s) \rightleftharpoons Ag + (aq) + Br - (aq)
the solubility product constant is
Ksp=7.7×10−13=[Ag+][Br−]Ksp = 7.7 \times 10^{ - 13} = [A{g^ + }][B{r^ - }]
=s⋅s=s2= s \cdot s = {s^2} .
Thus, we have that the solubility in pure water is:
s=Ksp=8.775×10−7M  Br−s = \sqrt {{K_{sp}}} = 8.775 \times 10^{- 7}M\;B{r^ - } or Ag+A{g^ + } .
In NaBr(aq)NaBr(aq) , what we have is a new initial concentration of Br−B{r^ - } , in which the concentration of Br−B{r^ - } is 1:11:1 with that of NaBrNaBr .
Notice how the equilibrium previously had Ksp=s⋅s=s2{K_{sp}} = s \cdot s = {s^2} . This one instead is:
Ksp=7.7×10−13=s′(0.01+s′){K_{sp}} = 7.7 \times 10^{- 13} = s'(0.01 + s')
We assume that s′s' is small compared to 0.01 as Ksp<<10−5{K_{sp}} < < {10_{ - 5}} . Thus, we have a simplified expression:
7.7×10−13≈0.01s′ ⇒s′=7.7×10−11  M \begin{gathered} 7.7 \times 10^{- 13} \approx 0.01s' \\\ \Rightarrow s' = 7.7 \times 10^{- 11}\;M \\\ \end{gathered}
The true answer would have been from solving
0=(s′)2+0.01s′−7.7×10−130 = {(s')^2} + 0.01s' - 7.7 \times 10^{- 13}
which would give an identical answer because the percent dissociation is much less than 5%5\% .
Therefore, the introduction of NaBrNaBr into the solution decreased the solubility of Br−B{r^ - } ions by a factor of ≈11396\approx 11396 .

Note :
Solubility product constants can be calculated, and used in a variety of applications. Calculating Ksp′s{K_{sp}}'s from solubility data Calculating the solubility of an ionic compound in pure water from its Ksp{K_{sp}} . Calculating the solubility of an ionic compound in a solution that contains a common ion.