Chemistry, 20.01.2020 10:31 ayoismeisalex
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb=6.3×10− 5?
Answers: 2
in this question acts as a weak base. As seen in the equation in the question, 
rather than 
when it dissolves in water. The concentration of ![[\text{OH}^{-}]](/tpl/images/0462/5453/3da65.png)
from pH:
,
.![[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0462/5453/5eaa9.png)
.![[(\text{CH}_3)_3\text{N}]_\text{initial}](/tpl/images/0462/5453/27976.png)
:![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](/tpl/images/0462/5453/44236.png)
![[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}](/tpl/images/0462/5453/e63b0.png)
.
was also produced. The solution started with a small amount of either species. As a result, ![[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0462/5453/587e5.png)
.![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}](/tpl/images/0462/5453/be689.png)
,![[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}](/tpl/images/0462/5453/90893.png)
,![[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0462/5453/76ffc.png)
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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20?...
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