Question

If 1 mole of Potash alum is completely dissolved in water, how many moles of ions will be furnished?

Answer 1

Dissolution of Potash alum in water:

• The chemical formula for Potash alum can be given as ${\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{.}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{.}\mathbf{24}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.
• When 1 mole of Potash alum is completely dissolved in water, it dissociates to give 2 moles of ${\mathrm{K}}^{+}$ ions, 2 moles of ${\mathrm{Al}}^{3+}$ ions, 4 moles of ${{\mathrm{SO}}_4}^{2-}$ ions and 24 moles of ${\mathrm{H}}_2\mathrm{O}$.
• The balenced chemical reaction can be written as:

$$\begin{gathered} {\mathrm{K}}_2{\mathrm{SO}}_4.{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.24{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+4{{\mathrm{SO}}_4}^{2-}\left(\mathrm{aq}\right)+24{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Potash}\mathrm{alum})(\mathrm{Potassium}\mathrm{ion})(\mathrm{Aluminium}\mathrm{ion})(\mathrm{Sulphate}\mathrm{ions})\left(\mathrm{Water}\right) \end{gathered}$$

• Total number of moles of ions furnished $=2{\mathrm{K}}^{+}+2{\mathrm{Al}}^{3+}+4{\mathrm{SO}}_4^{2-}=8\mathrm{moles}\mathrm{of}\mathrm{ions}$

Therefore, when 1 mole of Potash alum is completely dissolved in water, 8 moles of ions will be furnished.