If you started with 100 mL and ended with 125 mL, you must have added 25 mL of the 0.1M AgNO3. If the (Ag^+) = (Cl^-) at this point you must have titrated all of the Cl^- and you have a saturated solution of AgCl and none of the Ag2CrO4 has pptd. What's the concn of the Ag^+ and Cl^-. That must be
(Ag^+) = (Cl^-) = sqrt(Ksp) = approx 1.3E-5 but you should do it over to obtain a better answer.
To ppt Ag2CrO4, enough AgNO3 must be added to exceed the Ksp for Ag2CrO4.
Ksp = 9E-12 = (Ag^+)^2(CrO4^2-)
Substitute 1E-4 for CrO4^2- and solve for Ag^+ in mol/L. Determine mols in 0.125L then use M = mols/L. You know M and mols, solve for L and convert to mL. Usually that gives you the answer, BUT when you do this you are ignoring the Ag^+ present from the solubility of AgCl and it appears to me that approx 1.3E-5M can't be ignored so I would use the formula of
(1.26E-5 +x)^2(1E-4) = 9E-12 and solve for x to find M Ag^+ needed to ppt Ag2CrO4 then add the steps of M Ag^+ needed x 0.125L and then the L = mols/M step. That may turn out that you could have ignored the Ag^+ from AgCl but you'll get practice solving quadratic equation anyway.
Check my work. If you have a follow up I suggest you copy this and start a new post at the top of the page since this one is down several pages.
In a titration for the determination of the chloride ion, a solution of 0.1M agno3 is used against a solution, the volume of which is 100ml. At the completion of the titration, the residual concentrations of silver and chloride ions are equal and the total volume is now 125ml. What additional volume of the silver nitrate solution is needed to effect the precipitation of silver chromate (Ag2CrO4) if the molar concentration of the chromate ion is 1x10^-4.
ksp of AgCl- 1.6x10^-10
Ksp of Ag2CrO4- 9x10^-12
I actually have no idea how to do this question. Please at least give me a starting point.
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