A .052 solution of benzoic acid, C6H5COOH, Ka=6.3x10-5, is titrated with a strong base. What is the (H+) of?

Question

...the solution one-half way to the equivalence point?

Answer 1

At equivalence point 0.052 mole of base were used to react with 0.052 moles of benzoic acid. At this point the concentration of C6H5COO- is 0.052M

Halfway to the equivalent point, the concentration of [C6H5COOH] is equal to the concetration of [C6H5COO-], each is 1/2 the original concentration. So from the formula

pH = pKa + log[A-]/[HA],

if [A-] = [HA] the log term is 0,

so pH = pKa, and [H+] = Ka = 6.3*10^-5M