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Concentratin of Solution

  1. Dec 5, 2008 #1
    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. jcsd
  3. Dec 5, 2008 #2

    chemisttree

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    You probably meant this...

    Cr(NO3)3 + NaOH ---> Cr(OH)3 (s) + 3NaNO3 (aq)
     
  4. Dec 6, 2008 #3
    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.
     
  5. Dec 6, 2008 #4

    Borek

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    Staff: Mentor

    Look at the reaction equation now - what is the substance that precipitated and weights 2.06 g?
     
  6. Dec 7, 2008 #5
    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.
     
  7. Dec 8, 2008 #6

    Borek

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    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.
     
  8. Dec 8, 2008 #7
    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.
     
  9. Dec 8, 2008 #8

    Borek

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    Staff: Mentor

    Sorry, my mistake. You are OK this time.
     
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