Concentration of NaOH Solution: Calculation and Analysis

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Discussion Overview

The discussion revolves around the calculation of the concentration of an NaOH solution after it reacts with HCl and subsequently with chromium (III) nitrate, leading to the formation of a precipitate. Participants explore the stoichiometry of the reactions involved and the appropriate use of molar masses in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • A student presents a problem involving the addition of NaOH to HCl and the subsequent reaction with chromium (III) nitrate, seeking help with calculations.
  • Participants discuss the correct reaction equations, with one participant correcting the initial equation to reflect the formation of chromium (III) hydroxide as a solid precipitate.
  • Another participant calculates the moles of chromium (III) hydroxide formed from the precipitate's mass and attempts to derive the concentration of NaOH from this information.
  • Concerns are raised about the accuracy of molar masses used in calculations, particularly regarding chromium (III) nitrate and chromium (III) hydroxide.
  • One participant suggests that the stoichiometric ratios in the reactions must be carefully considered, leading to a revised calculation of moles of NaOH needed for neutralization.
  • There is a discussion about whether to multiply or divide based on the stoichiometric ratios, indicating some confusion over the correct approach to the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the concentration of NaOH, with some agreeing on the need for proper stoichiometric ratios while others challenge the calculations and assumptions made. The discussion remains unresolved regarding the final concentration of NaOH.

Contextual Notes

Participants note potential errors in molar mass calculations and the need to account for neutralization reactions, but these issues remain unresolved as the discussion progresses.

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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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