Question

A solution contains Al3+ and Co2+. The addition of 0.3628 L of 1.714 M NaOH results in the complete precipitation of the ions as Al(OH)3and Co(OH)2. The total mass of the precipitate is 21.96 g. Find the masses of Al3+and Co2+ in the solution.

Answer 1

A solution contains Al3+ and Co2+. The addition of 0.3628 L of 1.714 M NaOH results in the complete precipitation of the ions as Al(OH)3and Co(OH)2. The total mass of the precipitate is 21.96 g. Find the masses of Al3+and Co2+ in the solution.

Moles of NaOH ( or OH- added) : 1.714 mol/L * 0.3628 L = 0.6218 moles

we have : Al3+ (aq) + 3 OH-(aq) ==> Al(OH)3

Co2+ (aq) + 2 OH-(aq) ==> Co(OH)2

we have ,                  3* Al3+ moles + 2* Co2+ moles = 0.6218 moles         ...   1

we get : Al3+ moles = (0.6218 moles - 2* Co2+ moles   )/3            .........1a

and ,    Al3+ moles* 78 g/mol + Co2+ moles *92.95 g/mol = 21.96 g       ...........2

from 2 and 1a

(0.6218 moles - 2* Co2+ moles   )/3* 78 g/mol + Co2+ moles *92.95 g/mol = 21.96 g   

16.17 g - Co2+ moles* 26 g/mol + Co2+ moles *92.95 g/mol = 21.96 g   

Co2+ moles *66.95 g/mol = 5.79 g

Co2+ moles = 0.08648 moles ====> mass of Co2+ in solution ( Co atomic mass : 28.01 g/mol ) = 2.422 g

and on solving further ,

Al3+ moles = 0.1785 moles ====> mass of Al3+ in solution ( Al atomic mass : 26.98 g/mol ) = 4.816 g