Given 0.114 M HNO2 in water with a Ka for that acid of 4.6x104, what is...

Question

Question

Given 0.114 M HNO2 in water with a Ka for that acid of 4.6x104, what is the molarity of NO2 present in that solution? 7.2x10

Answer 1

Answer #1

HNO2 -------- H+ + NO2 0.114 M ом initially ом XM (0.114-X) M XM equilibrium Ka = [H*][NO2] / [HNO2] 4.6 x 10-4 = XP / [0.11

Answer 2

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