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?

The balance is

This implies that the response will essentially proceed to its conclusion.

This turns into a problem with limiting reactants.

To find the concentration of $\text{AlF}_6^{3-}$ after mixing, we need to consider the reaction between aluminum ion ($\text{Al}^{3+}$) and fluoride ion ($\text{F}^-$). The balanced chemical equation for the reaction is:

$\text{Al}^{3+} + 6\text{F}^- \rightarrow \text{AlF}_6^{3-}$

Given the volumes and concentrations of the ions, we can use the formula for dilution to find the final concentration:

$C_1V_1 = C_2V_2$

where $C_1$ and $V_1$ are the initial concentration and volume of the aluminum ion, and $C_2$ and $V_2$ are the final concentration and volume after mixing.

Substitute the given values:

$(0.1 \, \text{mol/L}) \times (25 \, \text{cm}^3) = C_2 \times (25 \, \text{cm}^3 + 25 \, \text{cm}^3)$

$(0.1 \, \text{mol/L}) \times (25 \, \text{cm}^3) = C_2 \times (50 \, \text{cm}^3)$

$C_2 = \frac{(0.1 \, \text{mol/L}) \times (25 \, \text{cm}^3)}{50 \, \text{cm}^3}$

$C_2 = 0.05 \, \text{mol/L}$

So, the concentration of $\text{AlF}_6^{3-}$ after mixing is $0.05 \, \text{mol/L}$.