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Sufficient strong acid is added to a solution containing Na2HPO4 to neutralize one-half of it. what will be the pH of this solu...Asked by ERAGON
Sufficient strong acid is added to a solution containing Na2HPO4 to neutralize one-half of it. what will be the pH of this solution? Explain please
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Answered by
DrBob222
pH = pKa2 + log (base/acid)
HPO4^2- + H^+ ==> H2PO4^-
Suppose you start with any amount of HPO4^2- (let's say 10 moles). If you add H^+ to neutralize 1/2 of it, then you will have 5 moles H2PO4^- formed and 5 moles HPO4^2- left un-neutralized.
Plug into the HH equation, first line, base = acid so that fraction is 1.0, log 1.0 is zero so pH = pKa2.
HPO4^2- + H^+ ==> H2PO4^-
Suppose you start with any amount of HPO4^2- (let's say 10 moles). If you add H^+ to neutralize 1/2 of it, then you will have 5 moles H2PO4^- formed and 5 moles HPO4^2- left un-neutralized.
Plug into the HH equation, first line, base = acid so that fraction is 1.0, log 1.0 is zero so pH = pKa2.
Answered by
MIKE
why pKa2??
Answered by
DrBob222
There are 3Ka values for H3PO4. k1 is for mixtures of H3PO4 and H2PO4^-; k2 is for mixtures of H2PO4^- and HPO4^2- (which is what is in the problem); k3 is for mixtures of HPO4^2- and PO4^3-
Answered by
Vincenzio Seeko
I disagree Senior Toth, I got 3.
Answered by
Vincenzo Cicco
Vincenzio is a fraud
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