The equilibrium is
#"Al"^"3+" + "6F"^"-" ⇌ "AlF"_6^(3-)#
#K_"eq" = 1.0 × 10^25#.
That means that the reaction will essentially go to completion.
This becomes a limiting reactant problem.
#"Moles of Al"^(3+) = 0.025 color(red)(cancel(color(black)("dm"^3 "Al"^(3+)))) × ("0.010 mol Al"^(3+))/(1 color(red)(cancel(color(black)("dm"^3 "Al"^(3+))))) = "0.000 25 mol Al"^"3+"#
#"Moles of AlF"_6^"3-" color(white)(l)"from Al"^(3+) = "0.000 25" color(red)(cancel(color(black)("mol Al"^(3+)))) × ("1 mol AlF"_6^"3-")/(1 color(red)(cancel(color(black)("mol Al"^(3+))))) = "0.000 25 mol AlF"_6^"3-"#
#"Moles of F"^"-" = 0.025 color(red)(cancel(color(black)("dm"^3 "F"^"-"))) × ("0.10 mol F"^"-")/(1 color(red)(cancel(color(black)("dm"^3 "F"^"-")))) = "0.0025 mol F"^"-"#
#"Moles of AlF"_6^"3-" color(white)(l)"from F"^"-" = 0.0025 color(red)(cancel(color(black)("mol F"^"-"))) × ("1 mol AlF"_6^"3-")/(6 color(red)(cancel(color(black)("mol F"^"-")))) = "0.000 433 mol AlF"_6^(3-)#
#"Al"^(3+)# gives the fewest moles of #"AlF"_6^"3-"#, so #"Al"^(3+)# is the limiting reactant.
Thus, when the reaction is complete, we have #"0.000 25 mol AlF"_6^"3-"# in #"50 cm"^3# of solution.
∴ #["AlF"_6^"3-"] ="0.000 25 mol"/("0.050 dm"^3) = "0.0050 mol/dm"^3#
