Answer:
pH = 1.95
Step-by-step explanation:
For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.
Therefore... the [H⁺] concentraion and pH are computed as follows...
[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M
pH = -log[H⁺] = -log(0.01118) = 1.95