A solution is $0.100 M$ in $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and $1.00 \times 1 0^{- 3} M$ in $NaOH$ . Find the pH, using the exact quadratic for the ionisation of $NH X_{3}$ (no approximation on $x$ in the numerator).

Answer: B

Let

x

be the

[OH X^{-}]

from

NH X_{3}

x (0.001 + x) / (0.100 - x) = 1.8 \times 1 0^{- 5}

. Solving the quadratic gives

x = 9.26 \times 1 0^{- 4} M

, so

[OH X^{-}] = 1.93 \times 1 0^{- 3} M

, pOH = 2.72, pH = 11.28.