Question

After graphing the titration of a weak acid, a student determines that the equivalence volume is 5.00 ml NaOH. Calculate the K, of the weak acid using the data table below. NaOH delivered pH (mL) 10.00 4.67 12.50 5.56 4.35 6.67 4.55 7.09 15.00 9.13 2.75 x 10-6 7.41x10-10 3.16 x 10-3 Not enough information is given to solve for ka

Answer 1

By definition of the equivalence point we can say that at equivalence point all the acid is just converted into salt. Now, similarly at the half of the equivalence point half of the acid will form salt and half will remain as acid. Now, 5.00 ml of NaOH is the equivalence point. Then the half equivalence point will be at (5.00 ÷ 2) = 2.50 ml of the same NaOH. We know weak acid and its its salt mixture forms a buffer solution.

Now, the pH of a buffer solution is given by the following equation,

$pH = pK_a + log_{10}(\frac{[salt]}{[acid]})$

Now, at the half equivalence point we will have [salt] = [acid].

Thus, at half equivalence point, pH = pK_a .

Now, at half equivalence point, that is, when 2.50 ml of NaOH is added then the pH is 5.56 (given data).

Therefore,

So,

.

Ka = 10-5.56 = 2.75 x 10-6

Note: We have assumed the acid to be a monoprotic acid.