Concentration of NaOH Solution: Calculation and Analysis

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The discussion focuses on calculating the concentration of an NaOH solution after it reacts with HCl and forms a precipitate of chromium (III) hydroxide. The student initially struggles with the stoichiometry and molar mass calculations but eventually determines that 2.06g of Cr(OH)3 corresponds to 0.02 moles, leading to a calculation of 0.06 moles of NaOH needed for the reaction. The neutralization with HCl is also considered, resulting in a total of 0.1 moles of NaOH in the solution. The final concentration of the NaOH solution is calculated to be 2 M. The discussion highlights the importance of correct stoichiometric ratios and molar mass in chemical calculations.
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Homework Statement


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.

Homework Equations


pH=14-pOH

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:
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gt000 said:
Cr(NO3)3 + HCL -> Cr(OH)3 + 3NaNO3

You probably meant this...

Cr(NO3)3 + NaOH ---> Cr(OH)3 (s) + 3NaNO3 (aq)
 
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.
 
Look at the reaction equation now - what is the substance that precipitated and weights 2.06 g?
 
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.
 
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.
 
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.
 
gt000 said:
Am I not suppose to use Chromium (III) Hydroxide's molar mass? Isn't that compound the one that has precipitated?

Sorry, my mistake. You are OK this time.
 

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