Example #1: Calculate the pH of a solution made by adding 2.70 g of lithium oxide (Li2O) to enough water to make 1.300 L of solution.
Solution:
1) The chemical reaction:
Li2O(s) + H2O(ℓ) ---> 2 Li+(aq) + 2OH¯(aq)
2) Determine moles of Li2O that dissolve:
2.70 g / 29.8814 g/mol = 0.0903572 mol
3) Determine moles of hydroxide produced:
Li2O : OH¯ molar ratio is 1:2For every one mole of Li2O that dissolves, two moles of hydroxide are produced.
(0.0903572 mol) (2) = 0.1807144 mol
4) Determine the molarity of the hydroxide:
0.1807144 mol / 1.300 L = 0.139011 M
5) Determine the pOH, then the pH:
pOH = −log 0.139011 = 0.85695pH = 14 − 0.85695 = 13.143 (to three sig figs)
Example #2: Calculate the pH when 0.850 g of BaO is dissolved in sufficient water to make 1.00 L of solution.
Solution:
1) When BaO "dissolves" in water, it actually reacts with the water:
BaO(s) + H2O(ℓ) ---> Ba2+(aq) + 2OH¯(aq)The key point will be that one mole of BaO produces two moles of hydroxide.
2) Determine the moles of BaO:
0.850 g / 153.329 g/mol = 0.005543635 mol
3) Determine moles of hydroxide in solution:
(0.005543635) (2) = 0.01108727 molThe contribution of hydroxide from the water is ignored.
4) Determine the molarity of the hydroxide:
0.01108727 mol / 1.00 L = 0.01108727 M
5) Determine the pOH, then the pH of the solution:
pOH = −log 0.01108727 = 1.955pH = 14 − 1.955 = 12.045
Example #3: Calculate the pH when 0.450 g of N2O5 is dissolved in 1.50 L of solution.
1) N2O5 reacts with the water as follows:
N2O5(ℓ) + H2O(ℓ) ---> 2HNO3(aq)
2) Moles of N2O5: 0.450 g / 108.009 g/mol = 0.00416632 mol
3) Determine moles of hydrogen ion in solution:
Nitric acid is a strong acid, so it ionizes 100%, releasing two moles of hydrogen ion for every mole of N2O5 that originally dissolved.(0.00416632 mol) (2) = 0.00833264 mol
4) Determine the molarity, then the pH:
0.00833264 mol / 1.50 L = 0.005555093 MpH = −log 0.005555093 = 2.255
Example #4: Chlorine(VII) oxide reacts with water to form perchloric acid. What is the pH when 0.380 g of Cl2O7 is dissolved into 1.00 L of solution?
Solution:
1) This is the relevant chemical equation:
Cl2O7(ℓ) + H2O(ℓ) ---> 2HClO4(aq)
2) Since perchloric acid is strong, it ionizes 100%:
Cl2O7(ℓ) + H2O(ℓ) ---> 2H+(aq) + 2ClO4¯(aq)
3) Determine moles of Cl2O7 that dissolve/react:
0.380 g / 182.901 g/mol = 0.0020776267 mol
4) Two moles of hydrogen ion are produced for every one mole of Cl2O7 that dissolves/reacts. Determine moles of H+ present in solution:
(0.0020776267 mol) (2) = 0.0041552534 mol
5) Determine the pH:
pH = −log 0.0041552534 = 2.381
Example #5: The solubility of CO2(g) in pure water is 0.0037 mol/L. Assuming that dissolved CO2 is in the form of H2CO3(aq), what is the pH of a 0.0037 M solution of dissolved CO2? Ka1 for H2CO3 = 4.3 x 10¯7
Solution:
1) The relevant chemical equation is:
H2CO3 ⇌ H+ + HCO3¯
2) Substituting into the Ka expression, we find:
4.3 x 10¯7 = [(x) (x)] / 0.0037x = 3.989 x 10¯5 M
3) Determine the pH:
pH = −log [H+] = −log 3.989 x 10¯5pH = 4.40 (to two sig figs)
Bonus Example: If 0.50 moles Ca(OH)2 is slurried in 0.50 L deionized water and treated with 0.50 moles of CO2 gas in a closed system, the liquid phase of this system will have a pH closest to what value?
Solution:
After the Ca(OH)2 and the CO2 react, we are left with some calcium carbonate, an insoluble substance. However, from the Ksp of CaCO3, we can calculate the approximate molarity of carbonate in the aqueous phase.I will use 5.5 x 10¯5 M for the carbonate concentration.
Carbonate is the salt of a weak acid and so it hydrolyzes in solution:
CO32¯ + H2O ⇌ HCO3¯ + OH¯
To describe that system, we require the Kb1 of carbonate, which we get from the Ka2 of carbonic acid, which is 4.7 x 10¯11.
So the Kb1 of carbonate is 2.13 x 10¯4 (from KaKb = Kw)
We can now calculate the [OH¯] in our calcium carbonate solution:
[OH¯] = SQRT[(2.13 x 10¯4) (5.5 x 10¯5)] = 0.000108 M
The pOH is just under 4, which makes the pH be just over 10.