I need (a second) math geek! Challenge for you

[toothybugs]

I have a fairly concentrated solution of sodium nitrate, and to manipulate the Redfield in my favor and partner with GFO, I would like to use it for nitrate dosing to help reduce PO4 in my struggling SPS-dom system (I measured PO4 at 0.5ppm the other day). But I would really like someone to confirm my numbers.

First, I took 50.0mL of solution and found it weighed 67.7g, or 1.354g/mL. I wanted to know the molarity or molality of such a solution so I could get a better ppm calculation.

Using this paper (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjVy7yv3PLMAhVCJR4KHTotCrMQFggcMAA&url=http://www.nrcresearchpress.com/doi/pdf/10.1139/v72-454&usg=AFQjCNF45zFJLLZu-0rWUZlLv82yK1FUHQ&cad=rja) and the densities and concentrations listed therein, I established a curve and found my solution to be approximately 7.07M. Given the mass of NO3 and NaNO3, I figure for my est'd 105 gallons of actual water, I need 3.5mL of my solution to adjust my system NO3 concentration by ~1ppm.

Am I right?

 

[jsker]

That is definitely a @Randy Holmes-Farley one

 

[Randy Holmes-Farley]

How did you have a nitrate solution that you did not know the concentration?

Yes, the 7 M part is roughly correct.

7M has 7 x 62 g/L = 434 g nitrate per L, or 434 mg/mL.

To boost 105 gallons (397 L) by 1 mg/L you need 397 L x 1 mg/L = 397 mg.

So just about 0.9 mL into your tank water would give ~ 1 mg/L. :)

 

[toothybugs]

That's what I get for overthinking a late-night chem problem :D Thanks Randy.

 

[JimWelsh]

I have a fairly concentrated solution of sodium nitrate, and to manipulate the Redfield in my favor and partner with GFO, I would like to use it for nitrate dosing to help reduce PO4 in my struggling SPS-dom system (I measured PO4 at 0.5ppm the other day). But I would really like someone to confirm my numbers.

First, I took 50.0mL of solution and found it weighed 67.7g, or 1.354g/mL. I wanted to know the molarity or molality of such a solution so I could get a better ppm calculation.

Using this paper (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjVy7yv3PLMAhVCJR4KHTotCrMQFggcMAA&url=http://www.nrcresearchpress.com/doi/pdf/10.1139/v72-454&usg=AFQjCNF45zFJLLZu-0rWUZlLv82yK1FUHQ&cad=rja) and the densities and concentrations listed therein, I established a curve and found my solution to be approximately 7.07M. Given the mass of NO3 and NaNO3, I figure for my est'd 105 gallons of actual water, I need 3.5mL of my solution to adjust my system NO3 concentration by ~1ppm.

Am I right?

Nice work on the density / concentration curve. What order polynomial did you use?

 

[tupes]

:eek: :confused: :confused: it's good to know we have smart people like these on the forum. Never knew it. Thanks for being a part of this to help folks out. Now I'm going take an advil after reading those equations.

 

[toothybugs]

Jim - I considered linear between the two surrounding points for simplicity in the determination but being a glutton for brain puzzles I opted for a 2nd order polynomial system, figuring the curve had to peak at some point (technically a horizontal asymptote, I guess) and we'd model the left side of it. I thought using a logarithmic curve would have been hellacious to work backward. In hindsight that may have been acceptable over the range of study as well.