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Reply. Hello again, Jason.
This is a math question. No, really it's an arithmetic problem. Your question is: if you have some
Aquarium Salt in the water of your 10 gallon Aquarium, and you remove 1 liter of that salty water each day, and replace that salty water with fresh tap water from your faucet, then how many days
will it take to get rid of all the salt?
The short answer is, "Never!" But 99% of the salt will be gone after 120 days. Lets do some arithmetic, and then finish with a few comments about changing water in
aquariums.
I like math and arithmetic. They sometimes help me take better care of my fish. You don't really have to do a lot of math or arithmetic to keep fish, but sometimes math helps,
and from time to time it helps a lot.
First, 1 liter = 1.05 quarts = 0.2625 gallons. So after you remove one liter of water from the 10 gallons of old salty water there will be 9.7375 gallons of water left in the
aquarium. Divide this by 10 gallons to get 0.97375 which is the ratio of the old salty water to total water in the aquarium. Or stated again using percentages: there is 97.375% old salty water
and 2.625% fresh water.
The old salty water and the new fresh water quickly mix together in your aquarium to produce slightly less salty water in your aquarium. The important number is the ratio
0.97375, because we can multiply this number by itself, 0.97375 x 0.97375, to get a result that is approximately 0.948189, and this number is the ratio of old salty water to the total amount of
water in your aquarium after two days.
To summarize, after the first day the ratio of old salty water to total water is about 0.97, and after two days the ratio is about 0.95. If we continue multiplying by 0.97375,
we get the following string of numbers: 0.923299, 0.8990642, 0.875462, 0.8524811, 0.8301034, 0.8083131, 0.7870948, and 0.7664335 after 10 days. We see the ratio of old salty water to total water
in your aquarium keeps getting smaller. As it should, as we remove a liter of the old salty water and add a liter of fresh water each day.
The last number, 0.7664335, is just the number we started with, 0.97375, raised to the 10th power. Or another way of saying the same thing, we've multiplied 0.97375 by itself a
total of 10 times to get approximately 0.7664335, which as I said before is about the ratio of old salty water to the total amount of water in your aquarium after 10 days.
If we multiply this number by itself, 0.7664335 x 0.7664335, we'll get 0.97375 to the 20th power. The answer is
0.7664335 x 0.7664335 = 0.5874202,
and this is the ratio of old salty water to total water after 20 days. Repeating this process
0.5874202 x 0.5874202 = 0.3450624 after 40 days,
0.3450624 x 0.3450624 = 0.1190680 after 80 days, 0.1190680 x 0.1190680 = 0.0141771 after 160 days.
So after 160 days the ratio is about 0.0141771 of old salty water to the total water. Restated using percentages: the water is about 1.42% percent old salty water and 98.58%
fresh water. Almost all of the salt is gone.
Can you see from this arithmetic that the ratio of salty water is never going to be zero? And that makes perfectly good sense, because when you remove a liter of water, you
always leave some water in the aquarium, and that water still contains some salt. The ratio is always getting smaller, but is never zero, so the amount of salt is always getting smaller but is
never quite zero.
So the answer to your question is: Never. That is to say, using your method of removing one liter of water each day and replacing it with fresh water, you will never get all
the salt out of your aquarium water. Of course, I know that you really didn't mean all the salt. You probably meant 99% or 98% of the salt, or something like that.
Maybe you also see that I've done a lot more arithmetic than I need to do, and I certainly would not have done all that arithmetic without a calculator. These long numbers
don't correspond to what's going on with the water, because neither you nor I can measure a liter of water to this degree of accuracy. I just wrote all these long numbers, because I was
concerned that if I rounded them all off you'd loose confidence in this calculation. Now you should have lots of confidence, but the trade off is that you are probably tired of seeing so many
long numbers.
I could have used 0.97 instead of 0.97373 and simplified my calculations by using just two figures after the decimal point.
0.97 x 0.97 = 0.94 after 2 days,
0.94 x 0.94 = 0.88 after 4 days, 0.88 x 0.88 = 0.77 after 8 days,
0.77 x 0.77 = 0.59 after 16 days, 0.59 x 0.59 = 0.35 after 32 days,
0.35 x 0.35 = 0.12 after 64 days, 0.12 x 0.12 = 0.01 after 128 days.
And the result is similar to the the first calculation where I used 0.97373 instead of 0.97. Now I see from doing both calculations that using 0.97 is accurate enough for me.
Good advice. Let me now switch from arithmetic to fish advice. It is not enough to change 3% of the water each day in your aquarium. You should change about 20% twice a
week. Now the calculation looks like this.
0.80 x 0.80 = 0.64 at the end of 1 week, 0.64 x 0.64 = 0.41 at the end of 2 weeks,
0.41 x 0.41 = 0.17 at the end of 4 weeks.
So by about the end of a month you will have replaced about 100% - 17% = 83% of the old water. Incidentally it is much better to change 20% of the water twice a week, rather
than 80% once a month, because changing more than 20% of the water on one day can easily shock your fish. Click
here to
read more about changing water in a warm water aquarium.
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