the following experiment was performed in the laboratory : an excess of solid SrCrO4 is vigorously shaken in water until a saturated solution is obtained. The excess solid is filtrated from the solution and 25.00 ml of the solution is titrated with 0.0115 M fe(no3)2 solution. A 30.60 ml portion of iron (ii) solution is needed to reach the endpoint. calculate the tsp of SrCro4. The titration reaction is CrO4^2- +3Fe^2+ +8H^+ ---- Cr^3+ +3Fe^3+ +4H2O

Question

the following experiment was performed in the laboratory : an excess of solid SrCrO4 is vigorously shaken in water until a saturated solution is obtained. The excess solid is filtrated from the solution and 25.00 ml of the solution is titrated with 0.0115 M fe(no3)2 solution. A 30.60 ml portion of iron (ii) solution is needed to reach the endpoint. calculate the tsp of SrCro4. The titration reaction is CrO4^2- +3Fe^2+ +8H^+ ----> Cr^3+ +3Fe^3+ +4H2O

DrBob222 · Answer

I have seen you write tsp twice. What in the world is tsp? Do you mean Ksp.
...........SrCrO4 --> Sr^2+ + CrO4^2-
I..........solid.......0........0
C..........solid.......x........x
E..........solid.......x........x

Ksp = (Sr^2+)(CrO4^2-)
You must determine the concentrations of Sr and CrO4. You do that with the Fe(NO3)2.
mols Fe(NO3)2 = M x L = ?
Using the titration equation, convert mols Fe(NO3)2 to mols CrO4^2-. That give you mols CrO4^2- in the 25.00 mL portion titrated. Convert that to mols in 1 L and that is (CrO4^2-). That is equal to (Sr^2+). Plug those into the Ksp expression and solve for Ksp.

Jack · Answer

Thanks!