Answered

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What is kb for (ch3)3N(aq)+H2O(l)= (CH3)3NH+(aq)+oh-(aq)?

Sagot :

Kb=[(Ch3)3 NH+][OH-] 
__________ 
[(Ch3)3 N]
that's the answer

Answer : The expression for the dissociation constant for weak base is :

[tex]K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]

Explanation :

[tex]K_b[/tex] is defined as the dissociation constant of a base. It is defined as the ratio of concentration of products to the concentration of reactants.

The chemical equation for the dissociation of weak base is :

[tex](CH_3)_3N(aq)+H_2O(l)\rightleftharpoons (CH_3)_3NH^+(aq)+OH^-(aq)[/tex]

The expression for dissociation constant, [tex]K_b[/tex] is given as:

[tex]K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]

The concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.  

Hence, the expression for the dissociation constant for weak base is :

[tex]K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]

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