Chemistry, 13.10.2019 21:30 nidiavega2009
You start with 0.050 moles of ammonia in 500. ml of water. the equilibrium constant keq is 1.8 × 10–5. what is the ph of this solution at equilibrium? show work!
Answers: 3
![\text{[B]} = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.050 mol}}{\text{0.500 L}} = \text{0.100 mol/L}](/tpl/images/0316/7651/145d4.png)
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}](/tpl/images/0316/7651/4767e.png)
![\dfrac{x^{2}}{0.100} = 1.8 \times 10^{-5}\\\\x^{2} = 0.100 \times 1.8 \times 10^{-5}\\\\x^{2} = 1.80 \times 10^{-6}\\\\x = \sqrt{1.80 \times 10^{-6}}\\\\x = \text{[OH]}^{-} = 1.34 \times 10^{-3} \text{ mol/L}](/tpl/images/0316/7651/c562e.png)
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.34 \times 10^{-3}) = 2.87\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.87 = \mathbf{11.13}\\\\\text{The pH of the solution at equilibrium is } \boxed{\mathbf{11.13}}](/tpl/images/0316/7651/2b390.png)
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You start with 0.050 moles of ammonia in 500. ml of water. the equilibrium constant keq is 1.8 × 10–...
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