A 0.2 molar solution of formic acid is 3.2%ionised its iinisation constant is
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Let HA = formic acid
HA ==> H+ + A-
Ka = [H][A]/[HA]
HA ===> H + A
I..0.2.......0.....0
C..-0.0064....+0.0064...+0.0064
E..0.2-.0064....0.0064....0.0064
Ka = (0.0064)(0.0064)/0.1936
Ka = 2.12x10^-4
sorry in the attachment there is a calculation mistake
your answer is 2.12×10^-4
HA ==> H+ + A-
Ka = [H][A]/[HA]
HA ===> H + A
I..0.2.......0.....0
C..-0.0064....+0.0064...+0.0064
E..0.2-.0064....0.0064....0.0064
Ka = (0.0064)(0.0064)/0.1936
Ka = 2.12x10^-4
sorry in the attachment there is a calculation mistake
your answer is 2.12×10^-4
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