A solution contains $0.100 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and $0.0500 M$ $NaOH$ . Solve the ammonia equilibrium exactly for the extra $OH X^{-}$ it supplies, then find the pH (two decimals; $K_{w} = 1 0^{- 14}$ ).

Numerical answer type, tolerance 0.01.

Answer: 12.7

Let

x

be the

OH X^{-}

from ammonia:

x (0.0500 + x) / (0.100 - x) = 1.8 \times 1 0^{- 5}

, so

x = 3.60 \times 1 0^{- 5} M

. Total

[OH X^{-}] = 0.0500 + x = 0.050036 M

, so

pOH = 1.301

and

pH = 12.70

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