Given a diprotic acid, H2A, with two ionization constants of Ka1=2.3*10^ - 4 and Ka2=3.8*10^-12, calculate the pH for a 0.142M solution of NaHA.

The pH for a 0.142M solution of NaHA is 1.0

Explanation:

Calculating the PH value based on the given data H2A

H2A with Ka1 = 2.3 * 10-4

The concentration of the acid is 0.180 M and the Ka1 is 2.3*10-4

Rewrite the chemical equation as,

H2A - -> H+1 + HA-1 with Ka1 = 2.3 * 10-4

Bythe formula:

Ka = (products) / (reactants):

2.3 * 10-4 = (x2) / (0.180)

x2 = 0.180 * 2.3 * 10-4

x = 0.414 * 10^-4 = [H+]

x = 4.14 * 10^-3 = [H+]

Since pH = - log of hydrogen ion concentration,

The pH for a 0.142M solution of NaHA is 1.0