Question

IONIZATION PROBLEM!!
calculate the original molarity solution of acetic acid that is 3.0% ionized. Ka= 1.8x10^-5
the answer is 1.9x10^-2 M
Can anyone please show how to get this answer?? My teacher has not taught us how to solve this kind of problem and idk why he assigned it -_- Any help will do. Please and thank you.

Answer 1

write the dissociation of acetic acid (or any weak acid):

HA <-> H+ + A- then, Ka=[H+][A-]/[HA]

dissociation is 3%, so [H+]=(0.03)*[HA]=[A-]

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for example if the concentration is 100, then dissociation is 100(0.03)
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Ka=((0.03)*[HA])^2/[HA] -> 1.8x10^-5=(0.03x)^2/x

x=[HA], so solve for x