Question

The equilibrium constant for the reaction in aqueous solution
$H_3{BO}_3+gylcerin\rightleftharpoons (H_3{BO}_3-gylcerin)$ is $0.90$.
How many moles of glycerin should be added per litre of $0.10M$ $H_3{BO}_3$ so that $80$% of the $H_3{BO}_3$ is converted to the boric-acid-glycerin complex?

• A. $4.44$
• B. $4.52$
• C. $3.6$
• D. $0.08$

Answer 1

Answer: (B)

$4.52$

$K_0=\frac{\left[complex\right]}{\left[H_{2}B{O}_3\right]\left[glycerin\right]}=0.90$

$\Rightarrow \frac{\left[complex\right]}{\left[H_{3}B{O}_3\right]}=\frac{80}{20}$

$\therefore \left[glycrin\right]=\frac{80}{20\times 0.9}=4.44M$

$H_{3}B{O}_3+glycerin\rightleftharpoons (H_{3}B{O}_3-glycerin)$

At eqn. $0.1-x$ $x$

$x=0.1\times \frac{80}{100}=0.08$ ; $a-x=4.44$

$\Rightarrow a=4.44+0.08=4.52M$ or Initial moles $=4.52$