(i) 0.183 M solution of H2A:
[HA-] = [H+] = (Ka1*[H2A])1/2 = (2.09*10-4 * 0.183)1/2
Therefore, [HA-] = 6*10-3 M (Or) 0.006 M
And [H2A] = 0.183 - (6*10-3)
Therefore, [H2A] = 0.177 M
pH = -Log[H+] = -Log(6.18*10-3)
Therefore, pH = 2.21
Now, [A2-] = (Ka2*[HA-])1/2 = (3.96*10-11 * 6*10-3)1/2
Therefore, [A2-] = 4.874*10-7 M
(ii) 0.183 M solution of NaHA:
[A2-] = [H+] = (Ka2*[NaHA])1/2 = (3.96*10-11 * 0.183)1/2
Therefore, [A2-] = 2.692*10-6 M
And [HA-] = 0.183 - (2.692*10-6)
Therefore, [HA-] ~ 0.183 M
This is an amphoteric salt.
Formula: pH = 1/2 (pKa1 + pKa2)
pKa1 = -Log(Ka1) = -Log(2.09*10-4) = 3.68
pKa2 = -Log(Ka2) = -Log(3.96*10-11) = 10.4
i.e. pH = 1/2 (3.68 + 10.4) = 1/2 (14.08)
Therefore, pH = 7.04
i.e. [H+] = 10-7.04 = 9.1*10-8 M
Now, [H+] = [A2-] - [H2A]
i.e. 9.1*10-8 = 3*10-6 - [H2A]
i.e. [H2A] = 2.692*10-6 - 9.1*10-8
Therefore, [H2A] = 2.601*10-6 M
(iii) 0.183 M solution of Na2A:
[HA-] = [OH-] = (Kb2*[Na2A])1/2 = (10-14/3.96*10-11 * 0.183)1/2
Therefore, [HA-] = 6.798*10-3 M
And [A2-] = 0.183 - (6.8*10-3)
Therefore, [A2-] ~ 0.176 M
pH = -Log[H+] = -Log(10-14/6.8*10-3)
Therefore, pH = 11.83
Now, [H2A] = (Kb1*[HA-])1/2 = (10-14/2.09*10-4 * 6.798*10-3)1/2
Therefore, [H2A] = 5.703*10-7 M
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