The equilibrium constant for the reaction: H 3 BO 3 + glycerine ⇌ ( H 3 BO 3 + glycerine complex ) is 0.90 . How much glycerine (in mol ) should be added to one litre of 0.10 M H 3 BO 3 so

Question

The equilibrium constant for the reaction: H 3 ​ BO 3 ​ + glycerine ⇌ ( H 3 ​ BO 3 ​ + glycerine complex ) is 0.90 . How much glycerine (in mol ) should be added to one litre of 0.10 M H 3 ​ BO 3 ​ solution, so that 60% of the H 3 ​ BO 3 ​ is converted to boric acid-glycerine complex?

TestHub · Accepted Answer

Initial moles of boric acid = 0.10 m o l (in 1 L ). 60% converts ⇒ complex formed = 0.06 m o l , remaining H₃BO₃ = 0.04 m o l . Let glycerine added = x m o l , so free glycerine at equilibrium = ( x − 0.06 ) . Equilibrium: H₃BO₃ + glycerine ⇌ complex K = [ H 3 ​ B O 3 ​ ] [ g l ycer in e ] [ co m pl e x ] ​ = 0.90 0.90 = ( 0.04 ) ( x − 0.06 ) 0.06 ​ Solving ⇒ x − 0.06 = 1.67 ⇒ x ≈ 1.73 m o l

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