# Chemistry

When 50.00 mL of an acid with a concentration of 0.214 M (for which pKa = 4.87) is titrated with 0.100 M NaOH, what is the pH at the equivalence point?

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1. The pH is determined by the hydrolysis of the salt at the equivalence point.
If we call the acid HA (the problem doesn't identify it) nor does it say it is monoprotic or not (but only one pKa suggests monoprotic) so the anion will be A^-. Volume of NaOH to reach the equivalence will be 0.214*50/0.1 = approx 107 mL but you should do a more accurate calculation. (A^-) at the equivalence point will be approx (0.214*50/157) = approx 0.07 (again you should redo all of the calculations). The hydrolysis equation ............A^- + HOH ==> HA + OH-
I..........0.07............0....0
C..........-x..............x....x
E.........0.07-x...........x....x

Kb for A^- = Kw/Ka for HA) = (x)(x)(0.07-x).
Solve for x = (OH^-) and convert to pH.

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