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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Darkmatter1986 on January 25, 2015, 07:04:16 PM

Title: Precipitation
Post by: Darkmatter1986 on January 25, 2015, 07:04:16 PM
Hey guys,

I hope this is the correct place to post this question, if not I apologize!

If someone can help me out by providing the necessary steps and or point me in the right direction on how to answer this question that would be a great help. Im a bit lost on where to even begin.


"Soluble phosphorus (P) in water is regarded as a nutrient. Typically, P in wastewater exists in the form of HPO42-. It should be removed from wastewater before it is discharged into natural water systems. In wastewater treatment plants, precipitation is a frequently used method to remove P from wastewater. In the precipitation, a coagulant (e.g. FeCl3) is added to transform soluble P to insoluble chemicals (precipitates), as shown below.

FeCl3 + HPO42-  = FePO4 (down arrow) + H+ + 3Cl-

These precipitates can be easily removed from water through gravity sedimentation. As a result, the effluent is P-free.
   If a wastewater has 8 mg/L HPO42-, what amount (mg/L) of FeCl3 should be added to completely remove it?"
Title: Re: Precipitation
Post by: Arkcon on January 25, 2015, 07:54:21 PM
Welcome, Darkmatter1986: to the Chemical Forums.  According to the Forum Rules{click} (http://www.chemicalforums.com/index.php?topic=65859.0), we want to see your attempt.  You may not have had a chance to read the rules yet, but we do want you to follow them.  For starters, you should try to setup a word problem like this.  Did you realize, the only pertinent part of the question you wrote is this:

FeCl3 + HPO42-  = FePO4 (down arrow) + H+ + 3Cl-

These precipitates can be easily removed from water through gravity sedimentation. As a result, the effluent is P-free.
   If a wastewater has 8 mg/L HPO42-, what amount (mg/L) of FeCl3 should be added to completely remove it?"

You've been given one reactant in mg/L, and you want another reactant in mol/L.  But how are they related to each other?  You have an equation, a balanced chemical equation, but what else do you think you'll need to relate them to each other?
Title: Re: Precipitation
Post by: Darkmatter1986 on January 25, 2015, 08:21:54 PM
I apologize for not having attempted the problem first. Part of my problem is that I'm not even sure where to start. I appreciate your help in getting me moving here.


I'm guessing you first need to convert everything to the same units?
Title: Re: Precipitation
Post by: Arkcon on January 25, 2015, 08:40:45 PM
And those units are?  What are the units (understood) in a balanced chemical reaction?  We don't actually reach into a bottle on a shelf, pick up one FeCl3 molecule, slap on one HPO42- molecule, and watch one iron phosphate molecule fall to our feet while a proton and a chloride anion fly over our heads, do we?
Title: Re: Precipitation
Post by: Darkmatter1986 on January 25, 2015, 08:45:51 PM
Would it be g/mol?
Title: Re: Precipitation
Post by: Darkmatter1986 on January 25, 2015, 11:06:30 PM
Ok I think I may have figured out the process. I ended up coming up with 12.96mg of FeCl3
Title: Re: Precipitation
Post by: billnotgatez on January 25, 2015, 11:34:03 PM
You might want to show us the steps you did to get your answer
Title: Re: Precipitation
Post by: Darkmatter1986 on January 25, 2015, 11:47:30 PM
8mg/L HPO4 = 0.008 g/L

0.008 g HPO4 = 0.00008 mol


HPO4 = 96g/mol
FeCl3 = 162 g/mol


so 0.00008 x 162 g/mol = 0.01296g

=12.96mg FeCl3 required
Title: Re: Precipitation
Post by: billnotgatez on January 26, 2015, 02:48:42 AM
Could you tell us how you derived this

Quote
0.008 g HPO4 = 0.00008 mol


Just want to make sure you understand the concept
Title: Re: Precipitation
Post by: Darkmatter1986 on January 26, 2015, 07:37:50 AM
Could you tell us how you derived this

Quote
0.008 g HPO4 = 0.00008 mol


Just want to make sure you understand the concept

No its fine, I actually wanted to make sure I got it right. :) Chemistry has never been my thing in the past, but I do find I'm enjoying it a bit more than I have before.
Title: Re: Precipitation
Post by: billnotgatez on January 26, 2015, 09:38:09 AM
OK so
Could you tell us how you derived this

Quote
0.008 g HPO4 = 0.00008 mol