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) A)1.12 B)3.64 C)12.32 D)6.20 E)7.21

Question

Quizplus · Accepted Answer

The equilibrium pH is determined by the first dissociation constant (Ka1) since it is the strongest. Using the formula pH = -log[H+], where [H+] is approximately equal to the square root of (Ka1 * initial concentration), we find the pH 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