# College Chemistry

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Calculate the pH at the equivalence point if 25.00 mL of 0.010M barbituric acid (HC4H3N2O3) is titrated with 0.020M NaOH. (Ka barbituric acid = 1.0 x 10^-5)

• College Chemistry -

If we call barituric acid, HB, the equation is
HB + NaOH ==> NaB + H2O, therefore, you can see that at the equivalence point we simply have the salt of barbituric acid so the pH will be determined by the hydrolysis of the salt.
You need to know the concn of the salt. Since the NaOH is twice the concn of the acid you know the volume of the base will be 12.5 mL so the concn of the salt at the equivalence point will be M = mols/L = 0.01 x (25/37.5) = 0.00667
........B^- + HOH ==> HB + OH^-
I...0.00667............0....0
C.......-x.............x....x
E...0.00667-x..........x....x

Kb for B^- = (Kw/Ka for HB) = (x)(x)/(0.00667-x)
Solve for x = (OH^-) and convert to pH.
Be careful with the algebra, you may need to use the quadratic equation since the concn of the salt is so small; i.e., 0.00667-x may not be equal to 0.00667