An excess of solid SrCrO4 is shaken in water until a saturated solution is produced. The solution is separated from the solid and a 25.00 mL aliquot is titrated with 0.0115 M Fe(NO3)2. Exactly 30.61 mL of the titrant was needed to reach the endpoint. Write the balanced equation for this redox titration, and determine the Ksp of SrCrO4.
Balanced redoc reaction : CrO4^-2(aq) + 3Fe^+2(aq) + 8H^-1(aq) --> Cr^+3(aq) + 3Fe^+3(aq) + 4H2O(l)
Moles of Fe(NO3)2 used for titration = 30.61 mL x 0.0115 M/1000
= 3.52x10^-4 moles
Each mole of CrO4^-2 in solution required 3 moles of Fe^+2
Moles of CrO4^-2 in 25 mL = (3.52x10^-4) / 3 = 1.17x10^-4 moles
Molarity of SrCrO4 = 1.17x10^-4 moles *1000/25 = 4.69*10^-3
Ksp = [Sr++][CrO4--] = (4.69x10^-3)^2 = 2.20x10^-5
Ksp = [Sr^2+][CrO4^2-] = 0.014*0.014 =1.96*10^-4
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