Question

The equilibrium constant for the given reaction is $0.90$.

$H_{3}B{O}_3+\text{glycerine}\rightleftharpoons H_{3}B{O}_3+\text{glycerine complex}$

The amount of glycerine should be added to one litre of $0.10M{H}_{3}B{O}_3$ solution, so that 60% of the $H_{3}B{O}_3$ is converted to boric acid-glycerine complex in $M$ is :

• A. 1.73
• B. 2.43
• C. 3.53
• D. None of these

Answer 1

Answer: (A)

1.73

$H_3+B{O}_3+glycerine\rightleftharpoons complex,K_c=0.90$

$0.1$ $a$ $0$ $.........initial$

$0.1-x$ $a-x$ $x$ $.........eqbm$

Where,

$x=\frac{60\times 0.1}{100}=0.06$

$K=\frac{\left[complex\right]}{\left[H_{3}B{O}_3\right]\left[glycerine\right]}=0.90$

If $60\%$ of $H_{3}B{O}_3$ gives complex then $\left[H_{3}B{O}_3\right]left=40\%$

Given,

$\frac{\left[complex\right]}{\left[H_{3}B{O}_3\right]}=\frac{60}{40}$

Thus,

$0.9=\frac{60}{40\left[glycerine\right]}$

or $\left[glycerine\right]$ or $(a-x)=\frac{60}{40\times 0.9}=1.67M$

$\therefore a=1.67+x=1.67+0.06=1.73M$