a 0.2 M Solutions of formic acid is 3.2% ionsed so what is its ionisation constant
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k aplha={C (alpha) ^2}/1-alpha. Here C=0.2M.aplha=3.2%.HCOOH=H++COOH-
K aplha=[COOH-][H+]/[HCOOH]
=(C aplha x C aplha) /C (1-aplha)
=(0.2 x 0.032)^2/(0.2 x 0.968)
=2.1 x 10^-4 is correct answer
K aplha=[COOH-][H+]/[HCOOH]
=(C aplha x C aplha) /C (1-aplha)
=(0.2 x 0.032)^2/(0.2 x 0.968)
=2.1 x 10^-4 is correct answer
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