What is the equilibrium pH of a 0.515 M solution of H3PO4 (aq)? (Ka1 = 7.5...
What is the equilibrium pH of a 0.515 M solution of H3PO4 (aq)? (Ka1 = 7.5 × 10–3 , Ka2 = 6.2 × 10–8 , Ka3 = 4.8 × 10–13) a. 1.23 b. 3.75 c. 12.32 d. 6.30 e. 7.21
Homework Answers
As Ka1 and Ka2 are very very small in comparison to Ka1 ,hence the concentration of hydronium ion comes from first dissociation.
[H+] = (Ka1×C)1/2
= (7.5×0.515×10-3)1/2
=6.2×10-3 M
(a) is correct option
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