A 2.0 L solution made by adding solid NaClO to water resulted in a basic equilibrium with a pH of 10.50. How many moles of NaClO were originally added? Ka of HClO is 3.0 × 10⁻⁷.

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) =

d: the voltage of the secondary coil is less than that of the primary coil.

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