# chemistry

A lab technician tests a 0.100 mol/L solution of propanoic acid and finds that its hydrogen ion concentration is 1.16 x 10^-3 mol/L. Calculate the percent ionization of propanoic acid in water

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1. Let's call propanoic acid HP.
HP ==> H^+ + P^-

Ka = (H^+)(P^-)/(HP)
If 0.1 M HP acutaly has 1.16 x 10^-3 (H^+), then
ionization = 1.16 x 10^-3/0.1 and you can convert that to percent by multiplying by 100.

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posted by DrBob222

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