Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
The pH of the ethanolamine solution has been 11.5.
The pH of the solution has been defined as the concentration of hydrogen ion in the solution.
The pH has been the difference of pOH from 14.
The ethanolamine has been a weak base. The OH concentration of ethanolamine has been given as:
[tex]\rm OH^-=\sqrt{Kb\;\times\;Molarity}[/tex]
The Kb of sample has been given as [tex]3.2\;\times\;10^-^5[/tex]
The molarity of sample has been, 0.3 M.
Substituting the values for hydroxide ion concentration:
[tex]\rm OH^-=\sqrt{3.2\;\times\;10^-^5\;\times\;0.3}\\OH^-=3.098\;\times\;10^-^3[/tex]
The pOH has been the negative logarithmic values of hydroxide ion concentration. The pOH has been given as:
[tex]\rm pOH=-log\;OH^-\\pOH=-log\;3.098\;\times\;10^-^3\\pOH=2.5[/tex]
The pOH of the sample has been 2.5.
The pH of ethanolamine sample has been given as:
[tex]\rm pH=14-pOH\\pH=14-2.5\\pH=11.5[/tex]
The pH of the ethanolamine solution has been 11.5.
For more information about pH of the solution, refer to the link:
https://brainly.com/question/4975103
