🤓 Based on our data, we think this question is relevant for Professor Goldsby's class at FSU.
We’re being asked to calculate the [HClO2] at equilibrium when 1.0 L of a 0.35 M solution of HClO2 is added 0.18 mol of NaClO.
Since HClO2 is a weak acid it will not dissociate completely in the aqueous solution and NaClO.
Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). Therefore, the equilibrium constants for the following reactions are as follows:
HClO2(aq) ⇌ H+(aq) + ClO2–(aq); Ka1 = 1.1 × 10–2 (can be found in books and the internet)
ClO- (aq) + H3O+(aq) ⇌ HClO(aq) + H2O (l); Ka = 2.9 x 10-8 (can be found in books and the internet)
The overall reaction is:
HClO2 (aq) + ClO- (aq) ⇌HClO (aq) + ClO2- (aq)
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.
The Ka expression for the reaction is:
Liquids are ignored in the Ka expression.
Note that each concentration is raised by the stoichiometric coefficient: [HClO2], [NaClO2], [NaClO] and [HClO] are raised to 1.
To 1.0 L of a 0.35 M solution of HClO2 is added 0.18 mol of NaClO.
Calculate the [HClO2] at equilibrium.