Suppose a 0.19M aqueous solution of oxalic acid (H2C2O4) is prepared. Calculate the equilibrium molarity of C2O4−2.
You'll find information on the properties of oxalic acid in the ALEKS Data resource.
Round your answer to 2 significant digits.
pKa1=1.25
pKa2=3.81

0.0034M = [C2O4²⁻

Explanation:

When H2C2O4 is in water, the equilibrium occurs as follows:

H2C2O4(aq) ⇄ HC2O4⁻(aq) + H⁺(aq)

The equilbrium constant, Ka1 (10^-pKa1 = 0.056234) is defined as:

Ka1 = 0.056234 = [HC2O4⁻] [H⁺] / [H2C2O4]

As both  HC2O4⁻ and H⁺ comes from the same equilibrium,  [HC2O4⁻] = [H⁺]. We can say:

[HC2O4⁻] = X

[H⁺] = X

[H2C2O4] = 0.19-X

Where X is reaction coordinate.

Replacing:

0.056234 = [X] [X] / [0.19-X]

0.010684 - 0.056234X = X²

0.010684 - 0.056234X - X² = 0

Solving for X:

X = -0.135. False solution. There are no negative concentrations.

X = 0.0790M. Right solution.

That means: [HC2O4⁻] = 0.0790M

Now, the HC2O4⁻ is in equilibrium as follows:

HC2O4⁻ ⇄ C2O4²⁻ + H⁺

The equilibrium constant of this reaction, Ka2, is:

Ka2 = 10^-3.81 = 1.5488x10⁻⁴ = [C2O4²⁻] [H⁺] / [HC2O4⁻]

Based on the same of the first equilibrium:

1.5488x10⁻⁴ = [X] [X] / [0.0790M - X]

1.2236x10⁻⁵ - 1.5488x10⁻⁴X - X² = 0

Solving for X:

X = -0.0036M

X = 0.0034M = [C2O4²⁻]

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.