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) =
The solution yielded a basic solution and 6.4 × 10^-11 moles of NaClO were originally added.
The NaOCl produces OCl- which reacts with water to yield HOCl as follows;
The concentration of HOCl is obtained from;
pH = -log[HClO]
[HClO] = Antilog(-pH)
[HClO] = Antilog (-10.50) = 3.16 × 10^-11M
HOCl(aq) ⇄ H^+(aq) + OCl-(aq)
I 3.16 × 10^-11 0 0
C -x +x +x
E 3.16 × 10^-11 - x x x
Ka = [ H^+] [ OCl-]/[HOCl]
Ka of HClO = 3.0 × 10⁻⁷
3.0 × 10⁻⁷ = x^2/3.16 × 10^-11 - x
3.0 × 10⁻⁷ (3.16 × 10^-11 - x) = x^2
9.48 × 10^-18 - 3.0 × 10⁻⁷x = x^2
x^2 + 3.0 × 10^-7x - 9.48 × 10^-18 = 0
x = 3.2 × 10^-11 M
Now;
Number of moles = concentration × volume
Number of moles = 3.2 × 10^-11 M × 2.0 L = 6.4 × 10^-11 moles of NaClO were originally added
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