calculate the pH of 0.53 M Ba(OH)2 and of 0.500M H2SO4

Answer 1

First, write the balanced equation for this reaction:

Ba(OH)₂ + H₂SO₄ ==> BaSO₄ + 2H₂O

Next, find moles of each reactant that's present:

Assuming 1 liter of solution

moles Ba(OH)₂ = 0.53

moles H₂SO₄ = 0.500

And finding the moles in excess, we see that there is more Ba(OH)₂ so it will be left over and pH will be >7.

Moles Ba(OH)₂ left over after reaction = 0.53 - 0.50 = 0.03 molesBa(OH)₂ = 0.03 Ba(OH)₂ left

Ba(OH)₂ ==> Ba2+ + 2OH-

[OH-] = 2 x 0.03 M = 0.06 M

pOH = -log 0.06 = 1.22

pH = 14 - pOH

pH = 12.78

If this is two separate questions, i.e. calculate the pH for 0.53 M Ba(OH)2 and for 0.500 M H2SO4, then the answer will be quite different.

(a) 0.53 M Ba(OH)₂ ==> 0.53 M Ba²⁺ + 1.06 M OH^-

pOH = -log 1.06 = -0.025

pH = 14 + pOH

pH = 14.03

(b) 0.500 M H₂SO₄ ==> 0.500 M H⁺ + 0.500 M HSO₄^-

If we assume that the ionization of HSO₄^- is small relative to the initial ionization, then we have 0.500 M H⁺

pH = -log 0.500

pH = 0.30

If you must include the 2nd ionization of HSO₄^- ==> H+ + SO₄^2-, look up the Ka₂ for H₂SO₄ and use it to solve for the [H⁺].