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# Sodium Carbonate Dilution

## Sodium Carbonate Dilution

(OP)
Hello, I have a tank(s)/pipe system with the following set up: I have one tank with 250 gallons of 5% sodium carbonate in solution. This tank is fed into another tank with water added to achieve a final sodium carbonate solution of  0.10% in soultion (by wieght).

My questions are:
1) How much water do I need to add to the second tank to cut the 5% solution to 0.10 % solution?
2) If the flow rate from the second tank is 90 gpm, how much water gpm would I need to add to 5% solution to reduce the sodium carbonate to 0.10 % by weight?

### RE: Sodium Carbonate Dilution

Using g (gram) as mass unit:

Solution #1: salt (5) + water (95) = 100
Solution #2: salt (5) + water (4,995) = 5000

Assuming density of solution #1 is 1.05 g/cc, its volume would be 95.24 cc.

Add (4995-95)= 4900 g = 4900 cc water to 100 g (= 95.24 cc of solution #1) to get 4900+95.24 = 4995.24 cc of  solution #2. The density of this solution would be about 1.0 g/cc.

Therefore for any volume flow rate of solution #1 add water in the ratio of 4900÷95.24 = 51.45 volumes of water, totalling 52.45 volumes of solution #2.

For 90 gpm the water addition would be 90×51.45÷52.45 = 88.284 gpm water to 1.716 gpm solution #1. In round figures 88.3 gpm water for every 90 gpm solution #2.

I truly hope the above is correct. Please check my assumptions and logic.

### RE: Sodium Carbonate Dilution

(OP)
Hi 25362,

Thank you for your detailed and accurate response.

Yes, your logic is correct but I want to validate my understanding with your concurrence.

1) So if I had a 250 gallon tank of 5% sodium carbonate in solution, I would need to add 250 * (51.45/52.45) = 245.23 gallons of water. The final volume would be 250 gallons + 245.23 gallons = 495.23 total gallons of 0.1 sodium carbonate in solution.

2) Does your final statement infer that I need to add 88.3 gpm of water to 90 gpm of 5% sodium carbonate in solution to arrive at 88.3+90 = 178.3 gpm of 0.1% sodium carbonate?

3) Last, why did you subtract 4995-95=4900 cc? Where did 52.45 come from?

### RE: Sodium Carbonate Dilution

Dracula, let's start from the last question backwards:

3) A 5% solution already contains 95 mass units of water and 5 mass units of salt. In a 0.1% solution 5 mass units of salt would need 4995 mass units of water to reach a total of 5000 mass units. Since the original solution (100 units) already contains 95 units of water you need to add the difference (=4900 units) to reach a toral of 5000 units.

Now, while water mass units = volume units, the original 100 mass units of the 5% solution correspond only to 95.24 volume units, the ratio between 4900 volume units of water to 95.24 volume units of the original 5% solution is 51.45.

This means that to 1 volume of the original solution you must add 51.45 volumes of water, reaching a total of 52.45 volume units.

2) No, it means that to reach 90 gpm you need to take about 1.7 gpm of the 5% solution plus about 88.3 gpm of water to keep the 51.45 ratio found above. The rounding of the figures gives a ratio of 51.9 instead of 51.45.

1) So, if you have 250 gal of a 5% solution you must add 250 × 51.45 volumes of water, namely 12862.5 gal of water to drop the concentration to 0.1%.

To make a rough check. To reduce a solurion from a concentration of 5% to 0.1% one needs to dilute it about 50 times (5÷0.1), right? Thus, we are on the right track.

I sincerely hope I made myself clear.

### RE: Sodium Carbonate Dilution

(OP)
Hi 25362,

Your rough check of 50:1 is exactly what I calculated when I first looked at the flows. This seemed very high and thats why I made my post.

3)Item is understood and correct.
2)Item is understood and correct.
1)Yes, again you have validated my first assumption.

Great work and thanks again for all you help!

### RE: Sodium Carbonate Dilution

Why does this sound like homework?

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