Question

1. Calculate the pH of the diprotic acid H2TeO4 (1.1 x10-3M). Ka1 = 2.1 x 10-8 and Ka2 = 6.5 x 10-12

Answer 1

If hydrosulfuric acid (H2A for H2TeO4) is dissolved in water the following two reactions occur: H2A (aq) = H+ + HA- K1, where K1 = 2.1*10^-8

HA- = H+ + A2- K2, where K2 = 6.5*10^-12

Consider the first reaction: [H2A] = 0.02 - x (here we can aproximate this by disregarding the "-x" which is very small in comparison because Ka1 = 2.1*10^-8 So [H2A] = 0.02 [H+] = [HA-] = x Ka1 = [H+] * [HA-] / [H2A] = x^2 / 0.02 = 2.1*10^-8

Then x = 2.05*10^-5

From the second reaction, we have [HS-] = 2.05*10^-5 - y = 2.05*10^-5 (because Ka2 is a very small number in comparison) [H+] = 2.05*10^-5 + y = 2.05*10^-5 [S2-] = y. Ka2 = [H+] * [S2-] / [HS-] = 2.05*10^-5 * y / 2.05*10^-5 = y

Then y = 6.5 * 10^-12

The total [H+] is x + y, then

[H+] = 2.05*10^-5 + 6.5*10^-12 = 0.00002050

(as you can see here the second reaction will not affect the final pH value, this is becaese Ka2 is very small in comparison with Ka1, you can disregard the second reaction in diprotic acids where Ka1 > 1000*Ka2).

pH = -log (0.00002050) = 4.69