posted by Henry on .
The ionisation of water plays an important role in the equilibria of aqueous solutions. Kw= 1.008 x 10-14 mol2 dm-6 at 298K.
(a) Calculate the pH of the following solutions at 298K showing all your working. Assume full dissociation in each case.
(i) 0.054 mol dm-3 aqueous HNO3
(ii) 2.0 mol dm-3 aqueous NaOH
(iii) 0.2 mol dm-3 aqueous H3PO4
(b) Calculate the concentration of H+(aq) and the concentration of OH-(aq) for the solutions with the following pH at 298K showing all your working.
(i) pH = 3.20
(ii) pH = 11.0
i and ii. pH = -log(H3O^+)
iii. H3PO4 + H2O==> H3O^+ + H2PO4^-
Write Ka expression for k1, solve for H3O^+ an convert to pH.
Same formula as in i and ii to solve for (H3O^+). Then (H3O^+)(OH^-) = 1E-14