Finding the initial concentration

Question

The solubility product of $\ce{AgBr}$ is $7.7 \cdot 10^{-13}\:\mathrm{mol^2/L^2}$. What was the initial concentration of $\ce{AgNO3}$ solution, if the precipitation of $\ce{AgBr}$ appears after the addition of $20\:\mathrm{mL}$ of a $0.001$ molar solution of $\ce{NaBr}$ to $500\:\mathrm{mL}$ of the $\ce{AgNO3}$ solution.

I got the solution as $0.054\:\mathrm{M}$. I'm confused with the procedure. This is what I have done.

1. Precipitate occurs at $K_{sp}=Q$ and $Q=[\ce{Ag+}][\ce{Br-}]$
2. $[\ce{Ag+}] = [\text{(Vol of $\ce{AgNO3}$)}\cdot\text{Molarity} ] / \text{Total Volume of the mixture}$
3. Similarly for $\ce{Br-}$
4. $[\ce{Ag+}][\ce{Br-}] = 2\cdot10^5$

The answer I got it $0.054\:\mathrm{M}$ ? Is that correct ?

Answer 1

This is a titration problem for quantitatively determining the concentration of a solution.

What reaction does occur?

$\ce{AgNO3 (aq) + NaBr (aq) <=> AgBr v + Na+ (aq) + NO3- (aq)}$
or essentially $\ce{Ag+ + Br- <=> AgBr v}$

Why is the solubility product important?

The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.

What can you tell about the equilibrium state at the point when the first precipitate falls?

The solubility product is matched, hence $\ce{[Ag^+][Br^-]}<K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$

What is the amount of bromine ions added to the solution?

$n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$

What can you tell about the concentrations in the final mixture?
First, what is the concentration of bromide ions in this mixture?

$V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$
$c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$

Second, what can you tell about the concentration of silver ions in the final mixture?

$c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$

What is the number of moles of silver ions in the final mixture?

$n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$

What is the initial concentration of the silver nitrate solution?

$c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$

Answer 2

The key is to get the concentration of bromide ions and use that value in the Solubility Equation as defined in step 1 to get $\ce{[Ag^+]}$:

$K_{sp} = [Br^-][Ag^+]$

The analysis and procedure is fine, except the product in step 4 it's a little bit big. Check the algebra reorder there. The answer I get is $2\cdot 10^{-8}~\mathrm{M}$. I would comment but I'm new in Chemistry Beta and can't do that. Hope that helps,

Answer 3

The way you posted your calculation is confusing. You should be clear on what you want in your statement.

First, find the number of moles of $Br^-$,

$\#\ moles\ Br^-=0.020L\cdot 0.001 M$

$\#\ moles\ Br^-=2\cdot 10^{-5} moles$

Now find the concentration of $Ag^+$ in the 520 mL solution,

$K_{sp}=[Ag^+][Br^-]$

$[Ag^+]=\frac {K_{sp}}{[Br^-]}$

$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{\frac {2\cdot 10^{-5} moles}{0.520L}}$

$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{3.84\cdot 10^{-5} moles/L}$

$[Ag^+]=2.00\cdot 10^{-8} moles/L$

Now find the concentration of $AgNO_3$ of the initial solution

$[Ag^+]=2.00\cdot 10^{-8} moles/L\cdot \frac {0.520 L}{0.500 L}$

$[Ag^+]=2.10\cdot 10^{-8} moles/L$

So the concentration of $AgNO_3$ of the initial solution is $2.10\cdot 10^{-8} moles/L$.