Question

Write a balanced equation for the reaction of$\mathbf{\text{Al}}{{\mathbf{\left(}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\right)}}_{\mathbf{\text{6}}}}^{\mathbf{\text{3 +}}}$in aqueous $\mathbf{KF}$.

Short answer

The balanced equation for the reaction of $\mathbf{\text{Al}}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\mathbf{\text{3}}}$in aqueous is KF given below. $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3 +}}{\text{+ 6F}}^{\text{-}}\rightleftharpoons {\text{AlF}}_{\text{6}}^{\text{3 -}}{\text{+ 6H}}_{\text{2}}\text{O(l)}$.

Step-by-step solution

Concept Introduction

A chemical equation is a representation of a chemical process in the form of components. A balanced chemical equation is one in which the number of atoms in all of the molecules is equal on both sides of the equation.

Writing the balanced equation

To overcome this problem, we must first point out that the coordination number for $\text{Al(III)}$is 6. Thus, when $\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3 +}}$reacts with , $\text{KF}$the maximum number of ${\mathrm{F}}^{-}$ coordinated to $\text{Al}\left(\text{III}\right)$ will be six.

$$\begin{gathered} \mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6^{3+}+6{\mathrm{F}}^{-}\rightleftharpoons {\mathrm{AlF}}_6^{3-}+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ {\mathrm{K}}_{\mathrm{f}}=\frac{\left[{\mathrm{AlF}}_6^{3-}\right]}{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6^{3+}\right]\times {\left[{\mathrm{F}}^{-}\right]}^6} \end{gathered}$$

Liquid and solid substances are not included in the K_f expression.

Therefore, the balanced equation is:

$\text{Al}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}^{\text{3 +}}{\text{+ 6F}}^{\text{-}}\rightleftharpoons {\text{AlF}}_{\text{6}}^{\text{3 -}}{\text{+ 6H}}_{\text{2}}\text{O(l)}$