Answer:
0.66 moles of NaClO were originally added
Explanation:
When NaClO is added to water, the equilibrium that occurs is:
NaClO(aq) + H₂O(l) ⇄ HClO(aq) + OH⁻(aq)
Where Kb is:
Kb = [HClO] [OH⁻] / [NaClO]
You can obtain Kb from Ka, thus:
Kb = Kw / Ka = 1x10⁻¹⁴ / 3.0x10⁻⁷
Kb = 3.33x10⁻⁸
As pH = 10.50;
pOH = 14 - 10.50 = 3.50
[OH⁻] = 10^{-3.50}
[OH⁻] = 3.16x10⁻⁴M
As OH⁻ and HClO comes from the same equilibrium, [OH⁻] = [HClO]
Replacing in Kb expression:
Kb = [HClO] [OH⁻] / [NaClO]
3.33x10⁻⁸ = [3.16x10⁻⁴] [3.16x10⁻⁴] / [NaClO]
[NaClO] = 0.333M
As there are 2.0L of NaClO solution, moles added were:
2.0L * (0.33moles / L) =
<h3>0.66 moles of NaClO were originally added</h3>
