Five Examples

**Example #1:** Calculate the pH of a solution made by adding 2.70 g of lithium oxide (Li_{2}O) to enough water to make 1.300 L of solution.

**Solution:**

1) The chemical reaction:

Li_{2}O(s) + H_{2}O(ℓ) ---> 2 Li^{+}(aq) + 2OH¯(aq)

2) Determine moles of Li_{2}O that dissolve:

2.70 g / 29.8814 g/mol = 0.0903572 mol

3) Determine moles of hydroxide produced:

Li_{2}O : OH¯ molar ratio is 1:2For every one mole of Li

_{2}O 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) + H_{2}O(ℓ) ---> Ba^{2+}(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 N_{2}O_{5} is dissolved in 1.50 L of solution.

1) N_{2}O_{5} reacts with the water as follows:

N_{2}O_{5}(ℓ) + H_{2}O(ℓ) ---> 2HNO_{3}(aq)

2) Moles of N_{2}O_{5}:
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 N_{2}O_{5}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 Cl_{2}O_{7} is dissolved into 1.00 L of solution?

**Solution:**

1) This is the relevant chemical equation:

Cl_{2}O_{7}(ℓ) + H_{2}O(ℓ) ---> 2HClO_{4}(aq)

2) Since perchloric acid is strong, it ionizes 100%:

Cl_{2}O_{7}(ℓ) + H_{2}O(ℓ) ---> 2H^{+}(aq) + 2ClO_{4}¯(aq)

3) Determine moles of Cl_{2}O_{7} 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 Cl_{2}O_{7} 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 CO_{2}(g) in pure water is 0.0037 mol/L. Assuming that dissolved CO_{2} is in the form of H_{2}CO_{3}(aq), what is the pH of a 0.0037 M solution of dissolved CO_{2}? K_{a1} for H_{2}CO_{3} = 4.3 x 10¯^{7}

**Solution:**

1) The relevant chemical equation is:

H_{2}CO_{3}⇌ H^{+}+ HCO_{3}¯

2) Substituting into the K_{a} 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¯^{5}pH = 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 CO_{2} 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 CO_{2}react, we are left with some calcium carbonate, an insoluble substance. However, from the K_{sp}of CaCO_{3}, 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:

CO

_{3}^{2}¯ + H_{2}O ⇌ HCO_{3}¯ + OH¯To describe that system, we require the K

_{b1}of carbonate, which we get from the K_{a2}of carbonic acid, which is 4.7 x 10¯^{11}.So the K

_{b1}of carbonate is 2.13 x 10¯^{4}(from K_{a}K_{b}= K_{w})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 MThe pOH is just under 4, which makes the pH be just over 10.