Question

0.77 grams of the solid oxide of an alkali metal is added to 55 mL of water. The metal oxide reacts with water to produce a basic solution. In order to completely neutralize this solution, 26 mL of 0.60 M HCl is added.

Based on this information, what alkali metal was used in the reaction?

Answer 1

26 mL x 0.60 mmol/mL = 15.6 mmol of H+ consumed by acid-base reaction between H+ and OH-. (2 sig dig).

Therefore, the reaction that generated OH- must have produced 15.6 mmol of OH-. All metal oxides are made with O-2 anion, so the reaction was:
\[{\rm O}^{2-}(aq) + {\rm H}_2{\rm O}(l) \rightarrow 2 {\rm OH}^-(aq)\]
From the stoichiometry to produce 15.6 mmol of OH- you need 15.6/2 = 7.80 mmol of O-2 (two sig dig).

The mass of 7.80 mmol of O-2 is:

7.80 mmol x 15.9994 mg/mmol = 124.795 mg (two sig dig).

Leaving (0.77 - 0.124795) = 0.645205 g (two sig dp) for the mass of metal M.

Now we have to consider possible formulas for the oxide:

Group 1A: M_2O. We had 2 x 7.80 mmol = 15.6 mmol M, and it weighed 0.645205 g. Molar mass = 0.645205 g / 0.0156 mol = 41. g/mol.

Could be potassium (K), molar mass = 39.1 g/mol.

Group 2A: MO. We had 7.80 mmol of M and it weighed 0.645205 g. Molar mass = 0.645205 g / 0.0078 mol = 83. g/mol.

Could be strontium (Sr), molar mass = 87.6 g/mol.

Group 3B: M_2O_3. We had 2/3 x 7.80 = 5.20 mmol M and it weighted 0.645205 g. Molar mass = 0.645205 g / 0.0052 mol = 1.2 x 10^2 g/mol.

No real good choices, the closest is lanthanum (La), molar mass = 138.9 g/mol.

Ah...but now I see the problem specified an alkali metal, Group 1A, so of course the answer must be potassium, and the formula of the oxide K2O.