What is the equilibrium pH of a 0.835 M solution of H₃PO₄(aq)? (K_a1 = 7.5 $\times$ 10^-3,K_a2 = 6.2 $\times$ 10^-8,K_a3 = 4.8 $\times$ 10^-13)

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Quizplus · Accepted Answer

The equilibrium pH is determined primarily by the first dissociation constant (Ka1) because it is significantly larger than Ka2 and Ka3. Using the formula for pH = -log[H+], where [H+] is approximately equal to the square root of (Ka1 * initial concentration), the pH is calculated to be around 1.12.

What Is the Equilibrium PH of a 0 ×\times× 10-3,Ka2 = 6

What Is the Equilibrium PH of a 0 $\times$ 10^-3,K_a2 = 6