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kendarriuskj1990
19.07.2021 •
Chemistry
Suppose a 0.034 M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4 You'll find information on the properties of sulfuric acid in the ALEKS Data resource. Round your answer to 2 significant digits.
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Ответ:
[SO4²⁻] = 0.015M
Explanation:
When H2SO4 is dissolved in water, HSO4- is produced in a direct reaction as follows:
H2SO4 → HSO4- + H+
As 1 mole of H2SO4 produce 1 mole of HSO4-, the molarity of HSO4- in this first reaction is 0.034M
Now, the HSO4- is in equilibrium with SO42- and H+ as follows:
HSO4⁻ ⇄ SO4²⁻ + H⁺
Where the equilibrium constant, K, is defined as:
K = 1.2x10⁻² = [SO4²⁻] [H⁺] / [HSO4⁻]
Where [] are the equilibrium concentrations of each species in the reaction.
The equilibrium concentrations are:
[SO4²⁻] = X
[H⁺] = X
[HSO4⁻] = 0.034M - X
Where X is reaction coordinate
Replacing:
1.2x10⁻² = [X] [X] / [0.034-X]
4.08x10⁻⁴ - 1.2x10⁻²X = X²
4.08x10⁻⁴ - 1.2x10⁻²X - X² = 0
Solving for X:
X = -0.027M. False solution, there are no negative concentrations.
X = 0.015M. Right solution.
That means the equilibrium molarity of SO4²⁻,
[SO4²⁻] = X
[SO4²⁻] = 0.015MОтвет:
we don't have the gizmo so we can't solve it I think
Explanation: