Ammonia solution has pH 10.5 at 298 K. With $K_{b} = 1.8 \times 1 0^{- 5}$ and neglecting $[OH X^{-}]$ against the total concentration, calculate its concentration $c$ as $c \times 1 0^{3}$ in $mol L^{- 1}$ , to two decimal places. Use $[OH X^{-}]^{2} = 1 0^{- 7}$ exactly.

Numerical answer type, tolerance 0.02.

Answer: 5.56

pOH

= 3.5

, so

[OH X^{-}]^{2} = 1 0^{- 7}

c = [OH X^{-}]^{2} / K_{b} = 1 0^{- 7} / (1.8 \times 1 0^{- 5}) = 5.5556 \times 1 0^{- 3} M

, so

c \times 1 0^{3} = 5.56