What is the pH of a 0.40 mol L^-1 H₂Se solution that has the stepwise dissociation constants K_a1 = 1.3 × 10^-4 and $K _ { \mathrm { a } 2 } = 1.0 \times 10 ^{- 11} ?$ A)2.14 B)3.89 C)4.28 D)5.57

Question

Quizplus · Accepted Answer

The pH is primarily determined by the first dissociation constant $K_{a1}$. Using the formula $pH = -\log[\sqrt{K_{a1} \cdot C}]$, where $C$ is the concentration, we find the pH to be approximately 2.14.

What Is the PH of a 0 Ka2=1.0×10−11?K _ { \mathrm { a } 2 } = 1.0 \times 10 ^{- 11} ?Ka2​=1.0×10−11?

What Is the PH of a 0 $K _ { \mathrm { a } 2 } = 1.0 \times 10 ^{- 11} ?$