# Problem: 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.

###### FREE Expert Solution

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

Given:

92% (328 ratings) ###### Problem Details

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.

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Based on our data, we think this problem is relevant for Professor Marzal's class at UCF.

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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.