Separate solid samples of sodium oxide, Na2O, and sulfur trioxide, SO3, are each added to an excess of distilled water at room temperature until no further reaction occurs. Which row correctly gives the approximate pH of the resulting solution in each case?

AS Chemistry 9701: pH of Sodium Oxide and Sulfur Trioxide in Water (MCQ)

Step 1: Sodium oxide with water

$\text{Na}_2\text{O}$ is a purely ionic solid containing discrete $\text{O}^{2-}$ ions. It reacts completely with water to form sodium hydroxide, which is fully soluble: $\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}$

Because NaOH is a strong alkali that dissociates completely and dissolves fully, the resulting solution is strongly alkaline, with a pH close to $14$ .

Step 2: Sulfur trioxide with water

$\text{SO}_3$ is a simple molecular oxide, but it reacts violently and completely with water to form sulfuric acid: $\text{SO}_3\text{(s)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)}$

$\text{H}_2\text{SO}_4$ is a strong acid that ionises essentially completely in water, so the resulting solution is strongly acidic, with a pH close to $0$ – $1$ .

Step 3: Match to the correct row

$\text{Na}_2\text{O}$ gives a strongly alkaline solution (pH $\approx 14$ ) and $\text{SO}_3$ gives a strongly acidic solution (pH $\approx 0$ – $1$ ), so option A is correct.

Why the other options are wrong

B: Reverses the two oxides. $\text{Na}_2\text{O}$ is the strongly alkaline one, not $\text{SO}_3$ .
C: Also reverses the two oxides. $\text{SO}_3$ is the strongly acidic one, not $\text{Na}_2\text{O}$ .
D: Both values are far too mild. $\text{Na}_2\text{O}$ is fully soluble and strongly alkaline (unlike MgO, which gives only a mildly alkaline pH $\approx 9$ because $\text{Mg(OH)}_2$ is sparingly soluble); $\text{SO}_3$ reacts completely with water and does not give a neutral solution.

Final answer

$\text{Na}_2\text{O}$ gives pH $\approx 14$ ; $\text{SO}_3$ gives pH $\approx 0$ – $1$ , option A.