The OH ^ - concentration in a mixture of 5 . 0mL of 0 . 0504M NH _...

\left[{NH}_{4}Cl\right] = \left[{NH}_{4}^{+}\right] = 0 . 0504MV = 5 . 0mL\left[{NH}_{3}\right] = 0 . 0210MV = 2 . 0mL$</p><p>${K}_{b} = \dfrac{\left[{NH}_{4}^{+}\right]\left[{OH}^{-}\right]}{\left[{NH}_{3}\right]}$</p><p>\left[{OH}^{-}\right] = \dfrac{{K}_{b} \times \left[{NH}_{3}\right]}{\left[{NH}_{4}^{+}\right]} = \dfrac{1 . 8 \times {10}^{- 5} \times 0 . 0210 \times 2}{0 . 0504 \times 5} = 3 \times {10}^{- 6}M$