Question

40.0mL of 0.270M Ba(OH)2 was mixed with 2.0 mL of 0.33M Al2(SO4)3. Each product was insoluable in water-2 precipitates form!

Calculate the mass of the precipitate, and the molarities of the ions remaining in the solution.

Answer 1

Volume of Ba(OH)2 solution, V1 = 40.0 mL = 40.0mL x (1L / 1000 mL) = 0.0400 L

Concentration of Ba(OH)2 solution, M1 = 0.270M

Hence moles of Ba(OH)2  mixed = M1xV1 = 0.270M x 0.0400L = 0.0108 mol Ba(OH)2

Volume of Al2(SO4)3 solution, V2 = 2.0 mL = 2.0mL x (1L / 1000 mL) = 0.0020 L

Concentration of Al2(SO4)3 solution, M2 = 0.33M

Hence moles of Al2(SO4)3 mixed = M2xV2 = 0.33M x 0.0020L = 0.00066 mol  Al2(SO4)3

When  Al2(SO4)3 reacts with Ba(OH)2, they form Al(OH)3 and BaSO4 ,both of which are insoluble in water. The balanced reaction is

Al2(SO4)3 + 3Ba(OH)2 ---- > 2Al(OH)3 + 3BaSO4

1 mol 3 mol 2 mol 3 mol

Now we need to find the limiting reactant.

In the above balanced reaction 1 mol of Al2(SO4)3 reacts with 3 mol of Ba(OH)2

Hence 0.00066 mol of Al2(SO4)3 that will react with the moles of Ba(OH)2

=  0.00066 mol Al2(SO4)3 x [3 mol Ba(OH)2 / 1 mol Al2(SO4)3]

= 0.00198 mol Ba(OH)2

Hence only 0.00198 mol of Ba(OH)2 reacted leaving the excess of Ba(OH)2 unreacted. Hence Al2(SO4)3 is the limiting reactant here and is exhausted completely.Hence the amount of products formed can be calculated from the moles of Al2(SO4)3.

Hence moles of Al(OH)3 precipitation formed = 0.00066 mol Al2(SO4)3 x [2 mol Al(OH)3 / 1 mol Al2(SO4)3]

= 0.00132 mol Al(OH)3

Molecular mass of Al(OH)3 = 78.0 g/mol

Hence mass of Al(OH)3 precipitation formed = 0.00132 mol Al(OH)3 x 78.0 g/mol = 0.103 g  Al(OH)3

Similarly moles of BaSO4 precipitation formed = 0.00066 mol Al2(SO4)3 x [3 mol BaSO4 / 1 mol Al2(SO4)3]

= 0.00198 mol BaSO4

Molecular mass of BaSO4 = 233.4 g/mol

Hence mass of BaSO4 precipitation formed = 0.00198 mol BaSO4 x 233.4 g/mol = 0.462 g BaSO4

$\therefore$ Total mass of precipitate formed

= mass of Al(OH)3 + mass of BaSO4

=  0.103 g + 0.462 g = 0.565 g (answer)

Now the ions remaining in the solution is the excess of Ba(OH)2 i.e Ba²⁺(aq) and OH^-(aq)

Excess of Ba(OH)2 remain unreacted = 0.0108 mol - 0.00198 mol = 0.00882 mol Ba(OH)2

Hence moles of Ba²⁺(aq) = moles of Ba(OH)2 remain untracted =  0.00882 mol Ba²⁺(aq)

Total volume = 42.0 mL = 0.0420 L

Hence molarity of Ba²⁺(aq) = 0.00882 mol Ba²⁺(aq) / 0.0420 L = 0.21 M (answer)

Similarly

moles of OH^-(aq) = 2 x moles of Ba(OH)2 remain untracted = 2x 0.00882 mol = 0.01764 mol OH^-(aq)

Total volume = 42.0 mL = 0.0420 L

Hence molarity of OH^-(aq) = 0.01764 mol OH^-(aq) / 0.0420 L = 0.42 M (answer)