What would be the charge balance if you add ${\mathbf{MgCl}}_{\mathbf{2}}$ to the solution and it dissociates into${\mathbf{Mg}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{Cl}}^{\mathbf{-}}$?

Charge balance:

The overall positive charges in solution equals the overall negative charges in solution.

${\mathit{n}}_{\mathbf{1}}\left[C_1\right]\mathbf{+}{\mathit{n}}_{\mathbf{2}}\left[C_2\right]\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}{\mathit{m}}_{\mathbf{1}}\left[A_1\right]\mathbf{+}{\mathit{m}}_{\mathbf{2}}\left[A_2\right]\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

$\left|A\right|$-concentration of a anion and-magnitude of the charge of the anion.

Species found in solution include:${\mathrm{H}}^{+},{\mathrm{OH}}^{-},{\mathrm{H}}_2\mathrm{O},{\mathrm{ClO}}_4^{-},\mathrm{Fe}{\left(CN\right)}_{{}^6}^{3-},{\mathrm{CN}}^{-},{\mathrm{Fe}}^{3+},M{g}^{2+},C{H}_{3}OH,HCN,N{H}_3,N{H}_{4}MgC{l}_2$ is added to the solution.

Charge balance:

$n_1\left[{\mathrm{C}}_1\right]+n_2\left[{\mathrm{C}}_2\right]+......=m_1\left[{\mathrm{A}}_1\right]+m_2\left[{\mathrm{A}}_2\right]+......$where,

-concentration of a cation and-charge of the cation.

-concentration of a anion and-magnitude of the charge of the anion.

Because these don't provide any charge, neutral species are disregarded.

On the basis of electro neutrality:$\left[{\mathrm{H}}^{+}\right]+3\left[{\mathrm{Fe}}^{3+}\right]+2\left[{\mathrm{Mg}}^{2+}\right]+\left[{\mathrm{NH}}_4^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{ClO}}_4^{-}\right]+3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{3-}\right]+\left[{\mathrm{CN}}^{-}\right]+\left[{\mathrm{Cl}}^{-}\right]$

The charge balancing equation for the given solution was determined using the concept of electroneutrality.