Answers: 3
![[NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M](/tpl/images/0718/8632/272cb.png)
![[NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M](/tpl/images/0718/8632/cd416.png)
![Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}](/tpl/images/0718/8632/bdcae.png)
![pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94](/tpl/images/0718/8632/c67ec.png)
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