Question

The equilibrium constant for the reaction in aqueous solution $H_{3}B{O}_3+$ glycerin $\rightleftharpoons$ complex, is 0.90. 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

	$H_{3}B{O}_3$ $+$	glycerine	$⟶$ complex
Initial	$0.10$	$0$	$0$
Equilibrium	$0.02$	$x$	$0.08$

The expression for the equilibrium constant is $K=\frac{\left[complex\right]}{\left[glycerine\right]\left[H_{3}B{O}_3\right]}=\frac{0.08}{x\times 0.02}=0.9$

Thus, $x=4.44$.