For $H X_{3} PO X_{4}$ : $K_{a 1} = 7.1 \times 1 0^{- 3}$ , $K_{a 2} = 6.3 \times 1 0^{- 8}$ , $K_{a 3} = 4.2 \times 1 0^{- 13}$ . In a $0.100 M$ solution of $H X_{3} PO X_{4}$ (no other species added), solve the first step as an exact quadratic, take $[H X^{+}] = [H X_{2} PO X_{4} X^{-}]$ from it, and find $[PO X_{4} X^{3 -}]$ in units of $1 0^{- 18} M$ (two decimals).

Numerical answer type, answer in x 10^-18 M, tolerance 0.01.

Answer: 1.13 x 10^-18 M

First step:

x^{2} / (0.100 - x) = 7.1 \times 1 0^{- 3}

gives

x = [H X^{+}] = [H X_{2} PO X_{4} X^{-}] = 0.02333 M

. Then

[HPO X_{4} X^{2 -}] \approx K_{a 2} = 6.3 \times 1 0^{- 8} M

. Finally

[PO X_{4} X^{3 -}] = K_{a 3} [HPO X_{4} X^{2 -}] / [H X^{+}] = 1.13 \times 1 0^{- 18} M