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⁺]

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

the right answer is c i just did the question.