Question

Calculate the number of milliliters of 0.469 M Ba(OH)₂ required to precipitate all of the Al³⁺ ions in 179 mL of 0.565 M AlCl₃ solution as Al(OH)₃. The equation for the reaction is:

2AlCl₃(aq) + 3Ba(OH)₂(aq) Al(OH)₃(s) + 3BaCl₂(aq)

Answer 1

From the eqation below,

  

3 mols of Ba(OH)2 is required to precipitate 2 mols of AlCl3.

We are given with 179mL of 0.565 M AlCl3 solution. We know than 0.565 molar means, one litre of the solution contains 0.565 mols of AlCl3.

Hence, 179mL of solution contains,       

So , from above reaction we know that 3 mols of Ba(OH)2 is required to precipitate 2 mols of AlCl3 .

So we need mols. That is, 0.152 mols of  Ba(OH)2.

We have, 0.469 M Ba(OH)2 Solution. From the equation of molarity we can find how many millilitres of the solution is needed to precipitate  179mL of 0.565 M AlCl3 solution.

  

  

So we need 324 mL of  0.469 M Ba(OH)2 Solution to  precipitate  179mL of 0.565 M AlCl3 solution.