Asked by John
What is the solubility of MnS, in grams per liter, in a buffer solution that is 0.120 M HC2H3O2 - 0.490 M NaC2H3O2? For MnS, Kspa=3x10^7.
Answers
Answered by
DrBob222
First determine the pH of the buffer solution.
pH = pKa + log(base)/(acid) and convert pH to H^+.
You then have
Ksp = (Mn^2+)(S^2-)
You have k1 and k2 for H2S as
(H^+)(HS^-)/(H2S) = k1
(H^+)(S^2-)/(HS^-) = k2
and (Mn^2+) = (S^2-) + (HS^-) + (H2S)
This is four unknowns and four equations but it isn't that bad. You can solve these four without assumptions. Most quant books show how to do this.
pH = pKa + log(base)/(acid) and convert pH to H^+.
You then have
Ksp = (Mn^2+)(S^2-)
You have k1 and k2 for H2S as
(H^+)(HS^-)/(H2S) = k1
(H^+)(S^2-)/(HS^-) = k2
and (Mn^2+) = (S^2-) + (HS^-) + (H2S)
This is four unknowns and four equations but it isn't that bad. You can solve these four without assumptions. Most quant books show how to do this.
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