Is Chemistry my problem?

Discussion in 'Water Chemistry' started by suzique, Oct 10, 2008.

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  1. suzique

    suzique Flamingo Tongue

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    I was trying to figure that out too. I did put a silverside in for the anemone, which it didn't eat. It was in the tank for about 4 hours. And for the rocks and sandbed...I did scrape the coralline off the glass below the sandline. Could that have done it? It's really hard when you have your first problem:confused: I REALLY don't want to loose anything!

    Sue
     
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  3. suzique

    suzique Flamingo Tongue

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    Have been looking at the Korilla? 2's and 3's. What do you think I should go with?
    Sue
     
  4. amcarrig

    amcarrig Super Moderator Staff Member

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    Could have. How deep is your sand and how much did you disturb it by scraping the coralline below the sandline?
     
  5. suzique

    suzique Flamingo Tongue

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    Sand is probably 3-4 inches deep, I just went down about an inch but just around the sides and front of the glass. I did take the BTA out to smell. It smells fine, just alot lighter than it was when I put it in the tank, also looks thinner. When I purchased it, it was located at the bottom of the tank. I got to thinking maybe my lights were more than it was used to, so when I put it back in the tank, I placed it near the sand and it grabbed a rock. Surprisingly, I think it looks a little better already:-/

    Sue
     
  6. amcarrig

    amcarrig Super Moderator Staff Member

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    How is your water testing out today?
     
  7. PharmrJohn

    PharmrJohn The Dude

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    I would go one K-2 and one K-3. Keep both your MJ 1200s. With your skimmer at about 200 your are looking at

    Koralia-2 = 600GPH
    Koralia-3 = 850GPH
    MJ 1200 = 295x2 = ~600GPH
    Skimmer = ~200GPH

    Total = 2250GPH divided by 46g = 49x

    Good number. No problems with flow.
     
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  9. suzique

    suzique Flamingo Tongue

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    I just tested: Temp. 78 PH 7.8 is down from from my last test it was 8.2 Nitrite 0 Nitrate 0 Ammonia 0 Phosphate 0 Calcium 440 down from 480 and dKH and KH I am still confused about. My test kit API does not have a formula to follow. It has a conversion chart #of drops=dKH=ppmKH So: according to that it took 6 drops=6dKH=107.4 If I divide the # of drops by 50 and multiply by 2.8 to get the dKH it would be 8.4 Right? OHHH! :confused: I'm Soo CONFUSED:angry: Ssfftt Ssfftt don't tell me old dogs can't learn new tricks 'cuz I'm a Katt!! LOL:LolLolLol

    Sue
     
  10. suzique

    suzique Flamingo Tongue

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    GOSH!!.......I really like you:) You've just taken me by the hand and helped me with so many things, THANK YOU;D I'm trying to read as many FAQ's as I can but it really makes a difference when your mentors reinforce your thoughts.

    Sue
     
  11. That Guy

    That Guy Aiptasia Anemone

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    The number of drops equals dkh so in your case....
    6 drops = 6 dkh

    to convert dkh to ppm you multiply dkh by 17.9...so in your case:
    6dkh x 17.9 = 107.4ppm

    To convert ppm to meq/L (milliequivalents/ liter) you divide ppm by 50 so in your case:
    107.4ppm / 50 = 2.15 meq/L

    Here is where you are getting confused. In order to go from meq/L to dkh is where you multiply by 2.8
    So in your case:
    2.15meq/L x 2.8 = 6dkh

    Anyway doing a lot more math i am finding your alk being really low due to really low magnesium. Id get a magnesium test kit and fill us in with the results of that test.
     
  12. That Guy

    That Guy Aiptasia Anemone

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    for anyone good at chemistry please correct me if i am wrong :)
    here are constants used.
    Ca - Equivalent Weight = 20.04
    Mg - Equivalent Weight = 12.15
    CO3 - Equivalent Weight = 30.004
    CaCO3 - Equivalent Weight = 50.045

    Given Info:
    Ca = 480ppm or 480 mg/L
    alk = 6 dkh or 107.4 ppm, or 2.15 meq/l

    So with the number you gave here is how i found you to have a low magnesium level.
    CaCO3 due to Ca is found by
    (480 x 50.045) / 20.04 = 1198.683mg/l
    so
    (1198.683mg/l) / 50.045 = 23.952 meq/L
    so
    (23.952 meq/l) / 50.045 = .4786 meq/l

    Since we found that 480 ppm of calcium will equal 23.952 meq/L we can than find Magnesium

    107.4 - 23.952 = 83.448

    so we can than find
    23.952 / 50.045 = . 4786
    and
    83.448 / 50,045 = 1.667
    so
    (.4786 + 1.667) x 50 = 107.3ppm

    So to find the estimated concentration of magnesium in the system you would take
    83.448 x 12.15 = 1014.31 ppm
     
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