really challenging chemistry problem?

Question

this is a type of problem that i'm having trouble with:

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

if you can figure out the answer and explain it a bit that would be great and i will offer you my respect and gratitude. i tried and got 1.932 M NaOH, but i think i did it wrong, so i would like to compare answers.

thanks for the help. that is what i did and i just used a slightly off molar mass by mistake. i think the person leading the review section i attended was wrong because he was rambling on about the precipitate being CrCl3 and doing some long complex process that confused everyone there.

Answer 1

the first reaction is

HCl + NaOH -> NaCl + H2O

NaOH must be in excess to form the precipitate in the next reaction:

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

molar mass of Cr(OH)3 is 103 g/mol, since the mass is given you can calculate that 0.02 mol were present, which reacted with 0.06 mol of NaOH (from the reaction stoichiometry).

the first reaction consumed 0.04 mol of NaOH (from the given concentration and volume of HCl).

so in total 0.04 + 0.06 = 0.1 mol of NaOH was present in 50 mL, which gives a concentration of 2.0 mol/L. you got pretty close but i'm not sure how to account for that small error - the molar mass perhaps?