Question

Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

Answer 1

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

$2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}$

Converting moles of boric acid to mass:

$0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}} = 6.183 g$

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.