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
To calculate the pH of a solution, you need to know the concentration of H+ ions in the solution. In this case, we can assume full dissociation, which means that all of the HNO3, NaOH, and H3PO4 compounds dissociate completely in water.
(a)
(i) For the solution of HNO3, we can assume full dissociation:
HNO3 → H+ + NO3-
Since the concentration of HNO3 is given as 0.054 mol dm-3, the concentration of H+ ions will also be 0.054 mol dm-3. To calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(0.054)
(ii) For the solution of NaOH, full dissociation produces hydroxide ions (OH-):
NaOH → Na+ + OH-
Since the concentration of NaOH is given as 2.0 mol dm-3, the concentration of OH- ions will also be 2.0 mol dm-3. To calculate the pH, we use the equation:
pOH = -log[OH-]
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH
pOH = -log(2.0)
(iii) For the solution of H3PO4, full dissociation produces three H+ ions:
H3PO4 → 3H+ + PO4^3-
Since the concentration of H3PO4 is given as 0.2 mol dm-3, the concentration of H+ ions will be 3 times that, so it will be 0.2 × 3 = 0.6 mol dm-3. To calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(0.6)
(b)
To calculate the concentrations of H+ and OH- ions for the given pH values, we use the equation:
pH = -log[H+]
(i) For pH = 3.20:
[H+] = 10^(-pH)
(ii) For pH = 11.0:
[H+] = 10^(-pH)
Since pH + pOH = 14, we can find the concentration of OH- ions using the equation:
[OH-] = 10^(-pOH)
Remember to plug in the values of pH and pOH to calculate [H+] and [OH-].
Note: In these calculations, we assumed that water is the solvent, and the temperature is 298K.