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I did this and the pH that I got was 1.57 (I even checked for significant digits) but apparantly that is not the right answer.

2007-10-22 11:50:11 · 1 answers · asked by alfonsocarnucci 2 in Science & Mathematics Chemistry

1 answers

Let [H+] = X
HNO2 <==> H+ + NO2-
Ka = 7.1x10^-4 = [H+]*[NO2-] /[HNO2] = X^2/(1 - X)
Thus: 7.1x10^-4(1 - X) = X^2
X^2 + 7.1x10^-4*X - 7.1x10^-4 = 0
Solve this quadratic equation. Take the positive solution:
X = 0.02629
pH = - log([H+]) = - log(X) = - log(0.0263) = 1.58

2007-10-23 03:22:27 · answer #1 · answered by Hahaha 7 · 0 0

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