Solve Ka: Percent Ionization of Formic Acid - 2.85%

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SUMMARY

The discussion focuses on calculating the concentration of formic acid required for 2.85% ionization, given a dissociation constant (Ka) of 1.80E-4. The equation used is X^2 / (0.0285 - X) = 1.8E-4, where X represents the concentration of ionized formic acid. The initial calculation yielded X = 0.00226, leading to a concentration of 0.0262M, which was identified as incorrect. Participants are encouraged to clarify the interpretation of 0.0285X in the context of the problem.

PREREQUISITES
  • Understanding of acid dissociation constants (Ka)
  • Basic algebra for solving quadratic equations
  • Knowledge of percent ionization calculations
  • Familiarity with the concept of equilibrium in chemical reactions
NEXT STEPS
  • Review the calculation of percent ionization in weak acids
  • Study the quadratic formula application in chemical equilibrium problems
  • Learn about the assumptions in weak acid dissociation
  • Explore the impact of concentration on ionization in weak acids
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Chemistry students, educators, and anyone involved in acid-base equilibrium calculations will benefit from this discussion.

George3
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Homework Statement


What concentration of formic acid would give an aqueous solution in which 2.85 percent of the formic acid molecules are ionized? Assume that Ka for formic acid is 1.80E-4 .


Homework Equations





The Attempt at a Solution



X^2 / .0285 - X = 1.8E-4
X = .00226
M = .0285 - .00226 = .0262M
This is wrong. Any ideas?
 
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Let X be the concentration of formic acid before dissociation. 0.0285X represents what?

Can you get there from here?
 

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