Question

A student mixes 50.0 mL of 1.00 M Ba(OH)₂ with 81.2 mL of 0.450 M H₂SO₄.

Calculate the mass of BaSO₄ formed.

Calculate the pH of the mixed solution.

Answer 1

The balanced equation for the titration reaction of with is given by-

Now, when we mix the solutions, the above reaction is taking place and as both and the sulfuric acid are reacting in a 1:1 ratio, we just have to calculate the number of moles of each of the reactants and calculate which of them would be in excess.

So, the number of moles of in 50 mL of 1 M =

and the number of moles of in 0.0812 L of 0.450 M =

.

As you can observe from the number of moles of each that, is in excess in the solution and the number of moles of   remaining in the mixture of solution after titration is = moles with a concentration of mixed solution =

.

Now, we can calculate the pH from the ionic product of water as below-

From the ionic product of water,

Hence, and pH= .

Now, in the above reaction, is the limiting reagent, so the number of moles of barium sulfate formed = the number of moles of that reacted. The number of moles of reacted was = 0.03654 moles. Hence, the number of moles of barium Sulfate formed = 0.03654 moles =   .