Question

The equilibrium constant for the reaction in aqueous solution
$H_{3}B{O}_3\left(aq\right)+glycerin\left(aq\right)\rightleftharpoons (H_{3}B{O}_3-glycerin)\left(aq\right)is0.90$.
How many moles of glycerin should be added per litre of $0.10M{H}_{3}B{O}_3$ so that $80\%$ of the $H_{3}B{O}_3$ is converted to the boric acid -glycerin complex?

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

Answer 1

The correct option is A $4.44$

We need $\frac{\left[complex\right]}{\left[H_{3}B{O}_3\right]}=\frac{80}{20}$
$K_c=\frac{\left[complex\right]}{\left[H_{3}B{O}_3\right]\left[glycerin\right]}=0.90$
$\therefore \left[glycerin\right]=\frac{80}{20\times 0.9}=4.44M$