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 ?

  • $\begingroup$ Would be correct if you used the number at step 4 -- the number you correctly wrote in step 1! Where did $2\cdot 10^5$ come from? $\endgroup$ Commented Jul 28, 2014 at 9:13

3 Answers 3


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?

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

  • $\begingroup$ Martin - Are you sure that it's not just $\ce{[Ag+]}[\ce{Br-}]=7.7\cdot 10^{-13}$ (mol/L)$^2$ at the moment the solution starts to get cloudy? $\endgroup$ Commented Jul 28, 2014 at 5:37
  • $\begingroup$ @SilvioLevy I am very sure that is true. I understood the question in the way that the concentration of silver nitrate is looked for, before the sodium bromide is added to this solution. $\endgroup$ Commented Jul 28, 2014 at 5:57
  • $\begingroup$ Yes, the question asks for the concentration before the addition of NaBr, but what I'm talking about the concentrations at the moment the solution turns. Why is $[\ce{Ag+}]=[\ce{Br^-}]$? To put it another way: your answer does not use the solubility product. If the molarities are the same at your "equivalence point", how does $\sim$0.00004 molar of bromide coexist *in solution* with $\sim$0.00004 molar of silver ion, just before the solution turns? That would mean $[\ce{Ag+}][\ce{Br-}]=1.6\cdot 10^{-9}\gg 7.7\cdot 10^{-13}$. (See also my reply to the comment you added to another answer.) $\endgroup$ Commented Jul 28, 2014 at 6:18
  • $\begingroup$ @SilvioLevy You are right, I was thinking about a titration with Mohr's method (there is no english wiki on this), but there you add an indicator to ensure you reached the equivalent point, which is not true in this case. I have to rework the answer or delete it all together. $\endgroup$ Commented Jul 28, 2014 at 6:43
  • 1
    $\begingroup$ Perfect answer now, but I've made a suggestion for clarity, by editing the 3rd answer directly. I think you have the reputation to see it and approve it you think it helps. $\endgroup$ Commented Jul 28, 2014 at 7:31

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,


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,


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

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

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

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

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 $2.10\cdot 10^{-8} moles/L$.

  • $\begingroup$ This answer is basically right, but it fails to take into account that the volume of solution grew from 0.5L to 0.52L. @LDC3, maybe you can fix it and then whoever downvoted it will reconsider? $\endgroup$ Commented Jul 28, 2014 at 7:34
  • $\begingroup$ @SilvioLevy The question states "What was the initial concentration of $AgNO_3$ solution?" I just didn't make that statement at the end. $\endgroup$
    – LDC3
    Commented Jul 28, 2014 at 13:25
  • $\begingroup$ @SilvioLevy I see what you're saying. I made a mistake in calculating the silver concentration. $\endgroup$
    – LDC3
    Commented Jul 28, 2014 at 13:29

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