The 7% Solution

[Pacapal]

OK all you Apistophiles, I have some water parameter questions for you..............

Using a 7% solution of well water to 93% R.O. water, I come up with the following:

Ph - 6.8
Gh - 4 degrees
Kh - 3 degrees

Does this look acceptable for most Apistos to spawn? Are any doubtful within these parameters?

(R.O. by itself tests out at Ph 6.2 and Kh of under 1 degree.)

Should I lower percentage of well water to 6% or even 5%? What is ideal bearing in mind that the tank will be as well planted as possible (without CO2). I do want to leave at least some buffering capability in the water don't I? Thanks in advance for your comments.

David

 

[tjudy]

Hi Pacapal... Here is a post of an equation that I use to reconsitute RO water to exactly what I want, regardless of what my tap parameters are. If you know what your tap water has to start with, then you can use it to make the water whatever you want it to be...

I have a little book written by Dr. Rolf Geisler back in 1963 (published by TFH) titled Aquarium Water Chemistry. Cool book. Not as out of date as you would think.

Here is a formula from the book that can be used to reconstitute RO water accurately (all numbers are ppm):

1) List the tap water hardness above the RO hardness:

300


0

2) Place the desired hardness between and to the right

300

50

0

3) Calculate the differences in the diagonals (300-50 = 250, etc.)

300 50

50

0 250

The differences show the relative amounts of water to be used to reach the desired TDS. The upper number (50 in this example) is the tap water, and the lower number is the RO water. Therefore the ratio is 250:50 RO:Tap. That can be reduced to 5:1.

If you wanted 6 gallons of 50ppm water using the above example, you would mix 5 gallons of RO with 1 gallon of tap.

To figure that out for any volume, divide the volume by the sum of the numbers in the ratio 5:1. (5 + 1 = 6). A 120 gallon tank, divided by 6, equals 20 gallons. To use the ratio for that volume of water (120 gallons), you would use 20 gallons x 5 = 100 gallons of RO; and 20 gallons x 1 = 20 gallons of tap.

So long as you know the TDS of the water you are going to use to reconstitute the RO with, this calculation will always work.

 

[Pacapal]

The differences show the relative amounts of water to be used to reach the desired TDS. The upper number (50 in this example) is the tap water, and the lower number is the RO water. Therefore the ratio is 250:50 RO:Tap. That can be reduced to 5:1.

If you wanted 6 gallons of 50ppm water using the above example, you would mix 5 gallons of RO with 1 gallon of tap.

To figure that out for any volume, divide the volume by the sum of the numbers in the ratio 5:1. (5 + 1 = 6). A 120 gallon tank, divided by 6, equals 20 gallons. To use the ratio for that volume of water (120 gallons), you would use 20 gallons x 5 = 100 gallons of RO; and 20 gallons x 1 = 20 gallons of tap.

[/quote]

OK, so if my well water is 670 pmm TDS and I want 50 ppm, I would have 670:50 or 67:5 or 13.4:1. That leaves me with
13.4 + 1= 14.4 and 128 oz. divided by 14.4 = 8.888.
And 8.888 13.4 = 119 oz. of R.O. water and 9 oz. of well water for every gallon of finished water. A 7.6% solution, correct?

But isn't buffering capacity in the Kh? What do I want that value to be? ANd are the parameters listed in my original post OK or do I need to adjust? BTW......thanks very much for posting that neat formula.....very cool........

David

 

[tjudy]

KH is a concentration of carbonate salts in the water. They will not show up on a TDS/EC meter, but can be diluted using the same formula. The hard part is going be if your tap concentrations are such that you cannot get to a specific hardness and a specific TDS....

If that is the case, then you should consider reconsitituting the RO with salts and carbonates rather than tap water. There are some commercially available products that will reconsistute RO for TDS without adding any carbonate hardness.

All in all though, I do not think that your water parameters are in bad shape.