For the aqueous AlF63− complex =Kf×6.921019 at 25°C.Suppose equal volumes of 0.0096M AlNO33 solution and 0.52M KF solution are mixed. Calculate the equilibrium molarity of aqueous Al3+ ion.Round your answer to 2 significant digits.

We have given Kf for [AlF6]3- = 6.92×1019 at 25°C Concentration of Al(NO3)3 = 0.0096 M Concentration of KF = 0.52 M And these solution have equal volume and we have to ask for the equilibrium molarity of aqueous Al3+Concentration of Al(NO3)3 = 0.0096 M Concentration of KF = 0.52 M The formation reaction for [AlF6]3- is as follow Al3+ + 6F- → [AlF6]3- Stoichiometry of reaction is 1 : 6 : 1 As these solutions are mixed in equal volume we can say that looking at a stoichiometry of the reaction 1 mole of Al3+ ≡ 6 mole of F- So 0.0096 mole of Al3+ ≡ X mole of F- On cross multiplying we get X mole of F- = (6 mole of F-)(0.0096 mole of Al3+)( 1 mole of Al3+)= 0.0576 mole of F- From this we can say Al3+ is a limiting reagent and the F- is in excess concentration So the remaining concentration of F- = 0.52 - 0.0576 = 0.4624 M Therefore Kf = [AlF6]3-[Al3+][F-]6= [0.0096][Al3+][0.4624]6 [Al3+] = [0.0096][6.92×1019][0.4624]6 = 1.4×10-20 M Therefore equilibrium concentration of [Al3+] = 1.4×10-20 M