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0.220 M HC3H5O2 ( use Ka = 1.30 x 10 -5 ) and 0.220 M NaC3H5O2 .

2007-03-16 06:25:20 · 2 answers · asked by simonkf2002 1 in Science & Mathematics Chemistry

2 answers

It sound like a buffer solution preparation:

1.- We use the Henderson Hasselbach equation:

pH = pKa + log [Acid]/[Base]

pH = -log(1.3x10^-5) + log (0.220/0.220)

pH = 4.88 + log(1) = 4.88 + 0

pH = 4.88

That's it!

Good luck!

2007-03-16 06:35:19 · answer #1 · answered by CHESSLARUS 7 · 0 0

pH = pKa = 4.886

2007-03-16 13:45:44 · answer #2 · answered by ag_iitkgp 7 · 0 0

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