# What is the concentration of "AlF"_6^"3-" after mixing "25 cm"^3 of 0.1 mol/L aluminium ion with "25 cm"^3 of 1 mol/L fluoride ion?

## For "AlF"_6^"3-", K_text(eq) = 1.0 × 10^25

Jul 8, 2016

I get ["AlF"_6^"3-"] ="0.0050 mol/dm"^3.

#### Explanation:

The equilibrium is

${\text{Al"^"3+" + "6F"^"-" ⇌ "AlF}}_{6}^{3 -}$

K_"eq" = 1.0 × 10^25.

That means that the reaction will essentially go to completion.

This becomes a limiting reactant problem.

$\text{Moles of Al"^(3+) = 0.025 color(red)(cancel(color(black)("dm"^3 "Al"^(3+)))) × ("0.010 mol Al"^(3+))/(1 color(red)(cancel(color(black)("dm"^3 "Al"^(3+))))) = "0.000 25 mol Al"^"3+}$

$\text{Moles of AlF"_6^"3-" color(white)(l)"from Al"^(3+) = "0.000 25" color(red)(cancel(color(black)("mol Al"^(3+)))) × ("1 mol AlF"_6^"3-")/(1 color(red)(cancel(color(black)("mol Al"^(3+))))) = "0.000 25 mol AlF"_6^"3-}$

$\text{Moles of F"^"-" = 0.025 color(red)(cancel(color(black)("dm"^3 "F"^"-"))) × ("0.10 mol F"^"-")/(1 color(red)(cancel(color(black)("dm"^3 "F"^"-")))) = "0.0025 mol F"^"-}$

${\text{Moles of AlF"_6^"3-" color(white)(l)"from F"^"-" = 0.0025 color(red)(cancel(color(black)("mol F"^"-"))) × ("1 mol AlF"_6^"3-")/(6 color(red)(cancel(color(black)("mol F"^"-")))) = "0.000 433 mol AlF}}_{6}^{3 -}$

${\text{Al}}^{3 +}$ gives the fewest moles of $\text{AlF"_6^"3-}$, so ${\text{Al}}^{3 +}$ is the limiting reactant.

Thus, when the reaction is complete, we have $\text{0.000 25 mol AlF"_6^"3-}$ in ${\text{50 cm}}^{3}$ of solution.

["AlF"_6^"3-"] ="0.000 25 mol"/("0.050 dm"^3) = "0.0050 mol/dm"^3