# Chemistry

What is the pH of a 0.050 mol/L sulfurous acid sample taken from an acid lake? The answer should come to 1.7 but I'm not sure how to get to it.

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1. .........H2SO3 ==> H^+ + HSO3^-
I.........0.05.....0......0
C..........-x......x......x
E.......0.05-x.....x......x

k1 = (H^+)(HSO3^-)/(H2SO3)
You should show k values when you post because most texts don't have the same number. I found k1 for H2SO3 in a text of mine that lists 1.72E-2
0.0172 = (x)(x)/(0.05-x)
Solve for x = (H^+), then pH = -log(H^+)
Look in your text, use that k1 for the above and you should be ok. I think you will need to solve the quadratic and not make simplifying assumptions.

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posted by DrBob222

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