Determine the pKb for the base B given that the equilibrium concentrations are [B]=1.5 M, [HB+]=0.87...
Determine the pKb for the base B given that the equilibrium concentrations are [B]=1.5 M, [HB+]=0.87 M, and [OH−]=0.75 M.
- Report your answer with two decimal places.
Homework Answers
Ans :-
Partial dissociation of BOH base is :
B (aq) + H2O (l) <-------------------------> HB+ (aq) + OH- (aq)
Expression of dissociation constant of base i.e. Kb (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).
Kb = [HB+].[OH-]/[B]
Kb = (0.87).(0.75) / (1.5)
Kb = 0.435
pKb of base = - log Kb
= - log 0.435
= 0.36
Therefore, pKb of base = 0.36
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