Calculate the pH of 4.9 ✕ 10−5 M boric acid (Ka = 5.8 ✕ 10−10).

Question

I do not know how to create the equation or use the Ka. Please help me. :(

Answer 1

Let boric acid be represented as HA. Then we can write the ionization of the boric acid as follows:

HA <==> H⁺ + A^-

Next, we write the Ka expression:

Ka = [H⁺][A^-] / [HA] and substituting values, we write...

Ka = 5.8x10^-10 = (x)(x) / 4.9x10^-5 - x (if we assume that x is small relative to 4.9x10-5, we can ignore it)

Ka = 5.8x10^-10 = (x)(x) / 4.9x10^-5

x² = 2.84x10^-14

x = 1.69x10^-7 = [H⁺] ... the above assumption was valid as this is only about 0.3% of 4.9x10^-5

pH = -log[H⁺] = -log 1.69x10^-7

pH = 6.77