The equilibrium constant for the reaction: H 3 BO 3 + glycerin ⇌ (H 3 BO 3 + glycerin complex) is 0.80. How many moles of glycerin should be added to 2 L of 0.20 M H 3 BO 3 solution, so that 50% of th

Question

The equilibrium constant for the reaction: H 3 BO 3 + glycerin ⇌ (H 3 BO 3 + glycerin complex) is 0.80. How many moles of glycerin should be added to 2 L of 0.20 M H 3 BO 3 solution, so that 50% of the H 3 BO 3 is converted to the boric acid-glycerin complex?

TestHub · Accepted Answer

At equilibrium [complex] = 0.10 M and [H₃BO₃] = 0.10 M 0.80 = 0.10/(0.10 × [glycerin]) gives [glycerin]_eq = 1.25 M

Glycerin consumed in the complex = 0.10 M, so total glycerin = 1.25 + 0.10 = 1.35 M

For 2 L: moles of glycerin = 1.35 × 2 = 2.70 mol

This is option (d).

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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