A stretegy for starting with very basic water

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jmills

jmills

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LifeTime Supporter
Jul 21, 2007
36
Billerica MA
Happy Spring,

I will be filling my pool again soon, so I am getting a fresh start on my water. With the advice of TFP I have been having progressively better results putting less in it each year. Visitors have commented on my pool water being very nice.

My tap water has a TA of 30. Best I can tell my tap water has a pH of 9.6. To measure this I mixed 2 parts of pH 7.2 distilled water with one part of my tap water. This tests as pH 8.

Assuming pH's average each other I computed my tap water pH as follows.

( pH + 7.2 + 7.2 )/3 = 8

pH + 14.4 + = 8 * 3

pH = 24 - 14.4 = 9.6

Adding the 72 ounces of baking soda to get my TA to 100 will probably add another 0.15 to the pH.

Adding the 23 ounces of CYA will probably drop the pH 0.75.

Last year I added muriatic acid to bring the pH down right away, but I wonder if I should bother at first. My water will drift more acidic over time.

What I am thinking of doing is adding the baking soda and CYA, maintain a fairly high chlorine level of about 8, but not worry about the high pH until the water is warm enough to swim in. With only 4.4k gallons of cold water the bleach bill will be very low.

When the temperature get to swimming temperature, about 67F, adjust the pH to 7.4.

Is this a good idea, or should I adjust the pH right away with the muriatic acid? One good point of dropping the pH now is I would be adding borax sooner and getting some borate in there.
 
Very simply add the water, let it circulate for several hours - and then test and go from there. Raise the TA first to 90-100 and allow that to circulate for and hour and retest both PH and TA. Then use the acid and target 7.4 on the PH. Let that circulate for an hour and retest again. I wouldn't wait to adjust these things and it doesn't take that long to make the adjustments. :)
 
pH doesn't average. If my rusty chemistry is correct, your water probably has a pH about 8.2. pH is the log of the hydrogen concentration of the solution, and is therefore not linear so the dilution method doesn't work.
 
At breakfast my wife told me she has a pH test kit for her fish tank that goes much higher. That test says the pH is greater than 8.6.

There is something about adding muriatic acid that bugs me, but I will get over it.

Thanks, JLM
 
While I advocate muratic acid as the pH lowering chem of choice we all have our monsters and if using it bothers you, you can use dry acid (sodium bisulfate) to lower your pH. Pool Calc even lists it. It does add a little sulfates to your pool but it works without having to handle liquid acid.
 
JohnT said:
pH doesn't average. If my rusty chemistry is correct, your water probably has a pH about 8.2. pH is the log of the hydrogen concentration of the solution, and is therefore not linear so the dilution method doesn't work.
Click to expand...

I didn't notice you had done a 2:1 dilution. The above calc is for a 1:1 dilution. For 2:1, it looks like your actual pH is 8.45. I'd let some sit in a glass for a few hours and see what the pH is after that.

If anybody can comment on the accuracy of my method, I used log(((10^7.2) + (10^7.2) + (10^x))/3)=pH, or the log of 1/3 of the sum of 10^pH of distilled plus 10^pH of distilled plus 10^pH of unknown. If I recall correctly, and it's been a long time, you add the total number of H+ atoms for the components to do the calculation.
 
Your formula is only correct due to a combination of errors that cancel each other out. Remember that pH=-log([H+]) or [H+]=10-pH, so the correct formula would have negatives in the exponents and on the logarithm, but this would only be correct when there is a significant excess of hydrogen ions to hydroxyl ions; that is, when the pH is significantly below 7.0, but that is not this situation. When there is a significant excess of hydroxyl ions to hydrogen ions then you add hydroxyl ions, 10(pH-14), instead of hydrogen ions, 10-pH. When the pH is closer to 7.0, then you can add both hydrogen ions and hydroxyl ions together separately, but then need to adjust to eliminate enough in pairs (to produce water) until you get to [H+]*[OH-] = 10-14.

There are also pH buffers in the water and normally that makes the computation much more complex if acid or base were added, but when adding distilled water this is a pure dilution effect so doing the simple calculation of cutting down hydroxyl ion to 1/3rd of its original value is sufficient. So the pH drops by log10(1/3) = 0.48 or about 0.5 units. Therefore, a diluted reading of 8.0 means an original reading of about 8.5. I'm ignoring the 7.2 of the distilled water and treating it as 7.0. Interestingly, your calculation seemed to work, but perhaps that just works out due to the double error canceling out (i.e. no negative on the exponents nor on the logarithm and not using hydroxyl ion so 14-pH).

Richard
 

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chem geek said:
Your formula is only correct due to a combination of errors that cancel each other out. Remember that pH=-log([H+]) or [H+]=10-pH,
Click to expand...

I knew the negaitives cancelled. I'm so used to working with logs (RF Power) that I just skipped ahead to the answer I knew I was going to get :roll: I also knew the answer was pretty well impossible if you started considering the other components you would be adding in tap water.

I'm curious why distilled water would have a pH of 7.2 though. Any insight?
 
JohnT said:
I'm curious why distilled water would have a pH of 7.2 though. Any insight?
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No idea except perhaps testing error. There is some temperature dependence on the pH of pure water, but to have a pH of 7.2 the water would need to be around 56-57ºF which I suppose could happen with tap water though that seems a bit cool. Certainly possible, especially if the pH were really more like 7.1 instead of 7.2.
 
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