<span>When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 </span> <span>pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) </span> <span>so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: </span> <span>pH = (7.2076+12.3767) / 2 </span> <span>pH = 9.79</span>
The salt K₂HPO₄ will dissociate to form the ions K⁺ and HPO₄⁻², and because of the stoichiometry of the dissociation (1:2:1) [HPO₄⁻²] = 0.600 M. So, two reactions may happen:
HPO₄⁻² + H₂O ⇄ H₂PO₄⁻ + OH⁻
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + OH⁻
H₃PO₄ is a polyprotic acid and have the values of pKa:
pKa1 = 2.15
pKa2 = 7.21
pKa3 = 12.4
In this case, the second and the third equilibrium are presented, thus, the pH depends only on the pKa values, and it will be:
