Consider a reaction ( HOCH2CH2)3 N + H2O (l) ( HOCH2CH2)3 NH + + OH -
For above reaction, dissociation constant K b = [( HOCH2CH2)3 NH +] [OH - ] / [ ( HOCH2CH2)3 N]
We can determine equilibrium concentrations of [( HOCH2CH2)3 NH +] & [OH - ] from pH.
We have , pH = -log [H + ]
[H + ] = 10 -pH = 10 - 10.655 = 2.21 10 -11 M
We have relation, [H + ] [OH - ] = 1.00 10 -14
[OH - ] = 1.00 10 -14 / 2.21 10 -11 = 4.52 10 -04 M = [( HOCH2CH2)3 NH +]
K b = ( 4.52 10 -04 ) ( 4.52 10 -04 ) / 0.329
Kb = 6.21 10 -07
In the laboratory, a general chemistry student measured the pH of a 0.329 M aqueous solution...
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