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hsedward hsedward
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3 years ago
50.0 mL of 0.360 M Cr(NO3)3(ag) is mixed with 100.0 mL of 0.270 M NaOH(ag). What is the mass of precipitate which forms? What are the concentrations of Cr?+ and OH after the precipitation process is
complete?
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wrote...
Educator
3 years ago
Hi hsedward

I think I figured out the first part:



Can you check to make sure it makes sense based on your understanding of the subject. I will look into your other question, and the second part of this a little later today.

Hope that helps
wrote...
Educator
3 years ago Edited: 3 years ago, bio_man
I believe to find the concentration of Cr2+ and OH-, use the chemical equation I provided on the left side:

Cr3+ + 3OH- Cr(OH)3

Given that the limiting reagent is NaOH, which produces a mass of Cr(OH)3 being 0.927 g, creating a mole ratio with Cr2+ and OH- gives them each a mole amount of 0.009 moles and 0.027 moles, respectively. To find their concentration, you divide this by total volume (150.0 mL) to get the moles per liter for each ion.

I believe that's how you do the second part of the question.

* If you have a sample question, please share it here so that I can double check my process from start to finish.
hsedward Author
wrote...
3 years ago
omg hey!! i don’t have any sample questions sorry but the Cr is supposed to be Cr3+ not 2+  and this question has choices and i chose the letter b but i’m not sure if it’s correct or not
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wrote...
Educator
3 years ago
That doesn't matter, just a typo Grinning Face with Smiling Eyes

I think you ought to pick d, look:

0.009 / 0.150 = 0.06
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