Calculate the ph of a 0.060 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11.

1) First dissociation

H2 CO3    =    H(+) + HCO3(-)          Ka1 = 4.3 * 10^ -7

0.06 - x              x          x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √[ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

2) Second dissociation

          HCO3(-)    =   H(+) + CO3(2-)         Ka2 = 5.6 * 10^ - 11

0.0001606 - y            y             y

Ka2 ≈ y^2 / 0.0001606 => y = √[0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

3) [H+] = x + y = 1.607 * 10^ -4

4) pH = - log [H+] = 3.79

3.79

Enzyme make a hormone as a catalyst