Calculate the [H+] in 1.0 M solution of Na2CO3 (for H2CO3, Ka1 = 4.3 × 10–7; Ka2 = 5.6 × 10–11). 7.5 × 10–6 M 1.3 × 10–2 M 7.5 × 10–13 M 6.6 × 10–4 M None of these choices are correct.

7.5x10⁻¹³M = [H⁺]

Explanation:

When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:

Na₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁺(aq) + OH⁻(aq) + 2Na⁺

Where Kb is defined from equilibrium concentrations of reactants, thus:

Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)

It is possible to obtain Kb value from Ka2 and Kw thus:

Kb = Kw /  Ka2

Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹

Kb = 1.8x10⁻⁴

Replacing in (1):

1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]

The equilibrium concentrations are:

[Na₂CO₃] = 1.0M - X

[HCO₃⁺] = X

[OH⁻] = X

Thus:

1.8x10⁻⁴ = [X][X] / [1-X]

1.8x10⁻⁴ -  1.8x10⁻⁴X = X²

X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0

Solving for X:

X = -0.0135 → False answer, there is no negative concentrations

X = 0.0133

As [OH⁻] = X;

[OH⁻] = 0.0133

From Kw:

Kw = [OH⁻] [H⁺]

1x10⁻¹⁴ = 0.0133[H⁺]

7.5x10⁻¹³M = [H⁺]

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

20 drops hope this