Question

A student added 50.0 mL of a sodium hydroxide solution to 100.0 mL of 0.400 M HCl. The resulting solution was then treated with an excess of aqueous chromium (III) nitrate, resulting in the formation of 2.06 g of precipitate. Determine the concentration of the initial sodium hydroxide solution.

Answer 1

Equations>

Cr(NO3)3 + 3NaOH --> Cr(OH)3 + 3NaNO3

-calculate the amount of moles of Cr(OH)

2.06 g Cr(OH)3 / 103.0 g/mol = 0.0200 mol

-calculate the amount of NaOH required to reaction completely with the Cr(OH)3

0.0200 mol Cr(OH)3 required 0.0600 mol NaOH (3x0.0200 mol)

You added 0.0400 mol HCl (0.100 L x 0.400 M = 0.0400 mol)
You neutralized 0.0400 mol NaOH but you still had 0.0600 mol left to form the precipitrate.
You started with 0.1000 mol NaOH (0.0400 + 0.0600 = 0.1000)
0.050 L NaOH x M = 0.1000 mol
M of NaOH = 2.000 mol/liter