Josh, did you proof your problem? You can calculate the original (acid) and (base) this way.
pH = pKa2 + log base/acid
7.2 = 7.2 + log base/acid
0 = log b/a
1 = b/a is equation 1
equation 2 is a + b = 0.05
Solve those two equations simultaneously and A = B = 0.025M
The buffer reaction is
HPO4^2- + H^+ ==> H2PO4^-
The final pH, of course, depends upon the H+ added but 0.06 M sounds unrealistic to me. Could that be 0.006M?
an ezyme -catalysed reaction was carried out in a solution buffered with 0.05 M phosphate ph 7.2 . As a result of the reaction ,0.06M of acid was formed. (phosphoric acid has three pka values. The one required for this calculation is
pka2=7.2).
1) what was the pH at the end of the reaction .
2) write the chemical equation showing how the phosphate buffer resisted a large charge in pH.
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