We're being asked to calculate the concentration of the NaOH solution.
First NaOH will react with NaOH:
NaOH(aq) + HCl(aq) → NaCl + H2O
Unreacted NaOH will react with chromium(III) nitrate (Cr(NO3)3) to form the precipitate:
3 NaOH(aq) + Cr(NO3)3(aq) → 3 NaNO3(aq) + Cr(OH)3(aq)
Step 1: Calculate the moles of NaOH reacted with HCl.
molarity HCl (volume) → moles HCl (mole-to-mole comparison) → moles NaOH
A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M 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.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Marzal's class at UCF.
What textbook is this problem found in?
Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.