Chemistry, 12.12.2019 03:31, ayoismeisjjjjuan
10. a 20.00 ml sample of 0.150 mol/l ammonia (nh3(aq)) is titrated to the equivalence point by 20.0 ml of a solution of 0.150 mol/l of the strong acid hydroiodic acid (hi (
a) write a balanced equation for the titration reaction.
b) what is the ph of the ammonia solution before the titration begins?
c) what is the ph at the equivalence point?
Answers: 2
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}](/tpl/images/0414/6924/2256c.png)
![\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}](/tpl/images/0414/6924/0b94c.png)
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}](/tpl/images/0414/6924/f4986.png)
![\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}](/tpl/images/0414/6924/607b2.png)
![\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}](/tpl/images/0414/6924/dc4d9.png)
![\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}](/tpl/images/0414/6924/b2d28.png)
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10. a 20.00 ml sample of 0.150 mol/l ammonia (nh3(aq)) is titrated to the equivalence point by 20.0...
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