I'll give you the question first and I'll show you what I came to....sooo....
A buffer solution is prepared by dissolving NaA and HA in pure water. The Sum of the molarities of NaA and Ha is 2.25 and the pH is 7.50. The Ka for HA is 3.0 x 10^-8. Calculate the molarities of NaA and HA in the solution.
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HA + H_2O <=> H_3O^+ + A^- (*NaH*)
pH = pka + log [ A^- / HA]
pka = log^(-Ka) => pka = log^(-3 x 10^-8)
pka= 1
7.50 = 1 + log [ A^- / HA]
6.50 = log [ A^- / HA] => 10^6.5 = [ A^- / HA]
[ A^- / HA] = 3.162 x 10^6
[A^-] + [HA] = 2.50
x + x = 2.50?
then I get lost in translation.... please help.
2007-08-03
15:00:21
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2 answers
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asked by
Catcher InTheRye
2
in
Science & Mathematics
➔ Chemistry