# Specific Gravity of Water

If I have a known volume of saline/saltwater and a known specific gravity, what is the formula I use to calculate how much salt water to replace to adjust the specific gravity?

Example:

I have a fish tank with 50 gallons of water with a specific gravity of 1.027 and I want to reduce the SG to 1.023 without changing the volume of water. How much water do I remove from the tank and replace with distilled water?

Yes, I know the real world answer is "some" and then retest... I just got curious last night as I was cleaning my fish tanks.

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 Sep 8, 2009, 11:46 AM
I guess that the easiest way is to find the concentration of the water in moles per gallon (I prefer moles per mL though, mL is the measure mostly used in my country when it concerns small volumes)

I've got a table of conversion of SG to ppt, and then from ppm to grams per kilogram.

1.027 : 38.1 ppt at 25 Celcius : 38.1 g/kg

But 1 kg of water is one litre; so 38.1 ppt = 38.1 g/L

Salt is NaCl, mass = 23 + 35.5 = 58.5 g/mol

Therefore, in one litre, we have 0.651 mol/L (concentration of salt) [38.1 / 58.5]

I'll take 5 Litres for the sake of units.

5 litres with concentration 0.651 mol/L

To be reduced to SG of 1.023 : 32.8 ppt : 0.561 mol/L [32.8 / 58.5]

Number of moles you have in 5 Litres of 1.027 = 0.651 * 5 = 3.26 mol

Number of moles you have in 5 Litres of 1.023 = 0.561 * 5 = 2.80 mol

You have to remove (3.26 - 2.80 = ) 0.46 mol of salt.

So, 0.46 mol is found in (0.46 / 0.651 = ) 0.71 L.

Therefore, in this case, you remove 0.71 L from your 5 Litres, and replace it with water to have the desired concentration

Sorry for being late, I got my puter on Sunday (from a week's repair) and had to look for the conversion tables for the SGs.
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #3 Sep 10, 2009, 10:20 AM
It's okay Steve. Just tell me when you need to change the salinity of your aquarium water.
 stevetcg Posts: 3,694, Reputation: 1824 Ultra Member #4 Nov 25, 2009, 05:47 AM
Quote:
 Originally Posted by Unknown008 It's okay Steve. Just tell me when you need to change the salinity of your aquarium water.
Ok... Now.

Currently we have 50 gallons @ 1.025 SG and need to reduce the specific gravity to 1.022. How much salt water do we replace with distilled?

Thanks!
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Nov 25, 2009, 06:49 AM
1. Ok, first basic assumption, I take the temperature as 23 celcius. (an average of the min and max in Florida.

2. 1.027 SG makes 37.5 ppt.

3. 50 gal -> 189.3 L

4. Concentration of salt is 0.641 mol/L

5. Number of moles in 50 gal = 0.641 * 189.3 = 121.3 mol

6. Desired concentration = 0.526 mol/L

7. Therefore, desired number of moles = 0.526 * 189.3 = 99.7 mol

8. Amount to be removed = 121.3-99.7 = 21.6 mol

9. Volume represented = 21.6/0.641 = 33.7 L

10. 33.7 L -> 8.90 gal

Therefore, remove 8.9 gal and replace it by water.

I just hope that my calculations are correct You can measure SG of your water right? Do a test first to see if that works.
 stevetcg Posts: 3,694, Reputation: 1824 Ultra Member #6 Nov 25, 2009, 07:00 AM
Quote:
 Originally Posted by Unknown008 1. Ok, first basic assumption, I take the temperature as 23 celcius. (an average of the min and max in Florida. 2. 1.027 SG makes 37.5 ppt. 3. 50 gal -> 189.3 L 4. Concentration of salt is 0.641 mol/L 5. Number of moles in 50 gal = 0.641 * 189.3 = 121.3 mol 6. Desired concentration = 0.526 mol/L 7. Therefore, desired number of moles = 0.526 * 189.3 = 99.7 mol 8. Amount to be removed = 121.3-99.7 = 21.6 mol 9. Volume represented = 21.6/0.641 = 33.7 L 10. 33.7 L -> 8.90 gal Therefore, remove 8.9 gal and replace it by water. I just hope that my calculations are correct You can measure SG of your water right? Do a test first to see if that works.
Youre the best. 2 points of adjustment (and I think I am beginning to understand too!)

Water temp is exactly 25 deg C and the current SG is 1.025.
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #7 Nov 25, 2009, 07:09 AM
Ok, ok , let's restart, at 25 C.

1. 1.025 SG makes 35.5 ppt.

2. 50 gal -> 189.3 L

3. Concentration of salt is 0.607 mol/L

4. Number of moles in 50 gal = 0.607 * 189.3 = 114.9 mol

5. Desired concentration = 0.578 mol/L

6. Therefore, desired number of moles = 0.578 * 189.3 = 101.9 mol

7. Amount to be removed = 114.9-101.9 = 13 mol

8. Volume represented = 13/0.0.607 = 21.4 L

9. 21.4 L -> 5.65 gal

Give it a try

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