Question

How much NAOH is required to prepare 50 mL of aqueous solution with 70 mg of Na+ ions per mL of the solution?

Answer 1

50 mL aqueous solution with concentration of Na⁺ ions 70mg/mL will contain:

$m(Na^+)=c(Na^+) \cdot V(Na^+) = 70\frac{mg}{mL} \cdot 50 mL = 3500 mg = 3.5 g$

The amount of Na⁺ ions in moles:

$n(Na^+) = \frac{m(Na^+) }{M(Na^+) } = \frac{3.5g}{23\frac{g}{mol}} = 0.1522 mol$

The same amount of NaOH is required to prepare solution:

$n(NaOH) = n(Na^+)=0.1522 mol$

Mass of NaOH is:

$m(NaOH) = n(NaOH) \cdot M(NaOH) = 0.1522 mol \cdot 40 \frac{g}{mol} =6.09 g$