Concentration of NaOH Solution: Calculation and Analysis

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SUMMARY

The discussion focuses on calculating the concentration of a sodium hydroxide (NaOH) solution after it reacts with hydrochloric acid (HCl) and chromium (III) nitrate. The student initially miscalculated the moles of NaOH based on the precipitate formed, chromium (III) hydroxide (Cr(OH)3), weighing 2.06g. The correct calculation involves recognizing the stoichiometric relationships in the reactions, leading to a final concentration of 2 M NaOH. Key equations include the neutralization reaction between NaOH and HCl, and the precipitation reaction with Cr(NO3)3.

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  • Understanding of stoichiometry in chemical reactions
  • Knowledge of molarity calculations
  • Familiarity with precipitation reactions and solubility rules
  • Ability to convert grams to moles using molar mass
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  • Study the concept of limiting reagents in chemical reactions
  • Explore molarity and dilution calculations in aqueous solutions
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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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