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.
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.
To solve this question, we need to consider the stoichiometry of the reaction and the solubility product constants (Ksp) of the reactants.
The reaction we are dealing with is:
AgCl + AgNO3 ⇌ Ag2CrO4 + NaNO3
First, let's determine the number of moles of Cl- ions present in the initial 100 mL solution. We know the initial concentration of AgNO3 is 0.1M, and the volume is 100 mL:
n(Cl-) = concentration(Cl-) x volume = 0.1 mol/L x 0.1 L = 0.01 mol
Since the residual concentrations of silver and chloride ions are equal, the concentration of residual Ag+ ions is also 0.01 mol. The volume is now 125 mL, so we can calculate the concentration of Ag+ ions:
concentration(Ag+) = n(Ag+)/volume(Ag+) = 0.01 mol/0.125 L = 0.08 M
Now, let's consider the solubility product constant (Ksp) of AgCl. The Ksp expression for AgCl is:
Ksp = [Ag+][Cl-]
Considering that [Ag+] = [Cl-] after the titration, we can simplify the expression:
Ksp = [Ag+]^2
Substituting our known values:
1.6x10^-10 = (0.08)^2
Solving for [Ag+]:
[Ag+] ≈ √(1.6x10^-10) = 4x10^-6 M
To precipitate silver chromate (Ag2CrO4), a reaction needs to occur when the concentration of Ag+ ions exceeds the Ksp value for Ag2CrO4. The Ksp value for Ag2CrO4 is 9x10^-12. Thus, we can set up and solve an equation to find the concentration of Ag+ ions required:
[Ag+]^2[CrO4-2] = Ksp for Ag2CrO4
Substituting the given Ksp value for Ag2CrO4 and the known [Ag+] value:
(4x10^-6)^2[CrO4-2] = 9x10^-12
Solving for [CrO4-2]:
[CrO4-2] ≈ 9x10^-12 / (4x10^-6)^2 ≈ 0.5625 M
The molar concentration of the chromate ion is given as 1x10^-4, which is higher than the calculated value of 0.5625 M. This means that the concentration of chromate ions is in excess, and the limiting reactant in this case is Ag+ ions.
To find the additional volume of silver nitrate solution needed, we can use the equation:
volume(AgNO3) = (n(AgNO3) / concentration(AgNO3)) - volume(initial solution)
Substituting the known values:
volume(AgNO3) = (0.01 mol / 0.08 mol/L) - 0.1 L = 0.125 L - 0.1 L = 0.025 L or 25 mL
Therefore, an additional volume of 25 mL of silver nitrate solution is needed to precipitate silver chromate (Ag2CrO4).