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Acid/Base Equilibria Problem 1

Incorrect!

Tutorial to help us answer problem 1.

 

To find the equilibrium concentration of H+, we can use the following equation,

[H+}{HC2O4^-]/[H2C2O4] = 5.6 x 10^-2.

We ignore the H+ contribution from the loss of the second proton because
Ka2 << Ka1. Let x denote [H+] at equilibrium. Then we have,

[H+][HC2O4-]/[H2C2O4] = x^2/(0.10 - x) = 5.6 x 10^-2.

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NOTE: If x is small compared to 0.10 (i.e. 0.10 − x = 0.10 to two significant figures), then we can neglect x. We can make this simplification when the value of [acid]/ Ka is large (i.e. greater than 100). We cannot make the simplification in this example, and we proceed.

We can now write the above equation as a quadratic equation by multiplying both sides by the denominator of the left hand side,

x^2 = (5.6 x 10^-2)(0.10 - x), x^2 = 5.6 x 10 ^ -2 (0.10 - x).

Expanding the above equation and bringing all terms to one side gives,

x2 + (5.6 × 10-2)x − (5.6 × 10-3) = 0.

The quadratic equation, ax2 + bx + c = 0, can be solved using the quadratic formula,

graphic with quadratic formula

yielding the solutions,

graphic solving above using quadratic formula

Since x represents the concentration of H+, the only answer that make sense is x ≈ 5.2 x 10-2 M.

 

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