# Homework Help: Concentratin of Solution

1. Dec 5, 2008

### gt000

1. The problem statement, all variables and given/known data
A student added 50.0mL 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 2.06g of precipitate. Determine the concentration of the NaOH solution.

2. Relevant equations
pH=14-pOH

3. The attempt at a solution
NaOH + HCL -> NaCL +H2O
Cr(NO3)3 + HCL -> Cr(OH)3 + 3NaNO3

I'm not sure what to do after creating the equations. I tried to use the 2.06g and converting this to moles using the molar weight of 3NaNO3, but i think I'm not doing the right steps. Any help?

Last edited: Dec 5, 2008
2. Dec 5, 2008

### chemisttree

You probably meant this...

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

3. Dec 6, 2008

### gt000

Yea I forgot to put the solid and aqueous. I'm still not really sure what to do, but I'll try it again and see if i get an answer.

4. Dec 6, 2008

### Staff: Mentor

Look at the reaction equation now - what is the substance that precipitated and weights 2.06 g?

5. Dec 7, 2008

### gt000

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

2.06g/103g = 0.02 mol Cr(OH)3
Cr(OH)3 * (1 mol NaOH/3 mol Cr(OH)3) = 0.0067 mol NaOH
Molary of NaOH =.0067 Mol/.05L = .133 M NaOH

I wrote the right equation this time. Here's my attempt, but I'm not sure if its correct.

6. Dec 8, 2008

### Staff: Mentor

You are partially right. First, your molar mass of chromium (III) nitrate is wrong. Second, you have calculated amount of excess NaOH that was left after it was partially neutralized with HCl, but you have neglected the neutralization.

7. Dec 8, 2008

### gt000

Am I not suppose to use Chromium (III) Hydroxide's molar mass? Isn't that compound the one that has precipitated?

Ok so another attempt:
After I obtained .02 ml Cr(OH)3, I multiplied by 3 instead of dividing because the ratio is 3:1 not 1:3.
0.02 x 3 = 0.06 moles NaOH
Since there needs to be neutralization with excess:
0.1L x 0.4 M = 0.04 moles HCl.
NaOH and HCl both have the same ratio of 1:1.
0.06 moles + 0.04 moles = 0.1 moles
0.1 moles/0.05 L = 2 M NaOH.

8. Dec 8, 2008

### Staff: Mentor

Sorry, my mistake. You are OK this time.