pH Chemistry Question

Question

If 7 mol of sulfuric acid (H₂SO_4(aq)) is used to create 8 L of solution, what is the pH of this solution?

Answer 1

7 mol H₂SO₄/8 L = 0.875 mol/L

Since you don't provide K2 for H₂SO₄, I'll assume that you are considering full dissociation of H₂SO₄, in which case you'd have 0.875 mol/L x 2 = 1.75 M H⁺

pH = -log [H+] = -log 1.75 = 0.24

If you only consider the 1st ionization, then you have 0.875 M H+ and

pH = -log 0.875 = 0.06

If you consider both ionizations, then you have...

H₂SO₄ --> H⁺ + HSO₄^-

HSO₄- --> H⁺ + SO₄^2- Ka₂

You'd need to incorporate the K₂ = [H⁺][SO₄^2-]/[HSO₄^-] and solve for H⁺, add it to the 0.875 and take the negative log.