4
$\begingroup$

I was reading an article that suggested that if one lets water sit for a day in a bottle or pot, then some of the metals in the water, such as lead, will settle to the bottom.

Does anyone know if this theory has been tested?

The reasoning seems somewhat logical, since particles in water tend to sink to the bottom. However, at the molecular level I know that ionic interactions may cause unintuitive behaviors.

I do have a TDS and pH meter. Perhaps I could let water sit for a day, then check the TDS of the water on top. Then compare that with the TDS of water that has been thoroughly mixed. Does anyone see any flaws with this idea?

Excessive toxic metals.

Many city water supplies, especially wells, contain some excessive amounts toxic metals or toxic amounts or forms of other minerals. Common toxic metals include arsenic and lead, which are found in many pesticides and herbicides and find their way into the wells and other water supplies. At times, these are difficult to remove from large amounts of water, so they are just left there even if the water violates environmental standards, especially in poorer nations.

Other minerals that often contaminate drinking water include too much iron, manganese, boron and copper depending on the location. One way to remove some of these easily is to let a pitcher or pot full of tap water sit for a day and some of the metals will settle to the bottom of the pot.

You can then pour off the water, leaving most of the minerals in the bottom of the pot. You may lose some beneficial minerals this way, but it is one way to remove some toxic metals, as well.

-Dr. Lawrence Wilson

http://www.drlwilson.com/ARTICLES/WATER.htm#INTRO

$\endgroup$
  • 2
    $\begingroup$ There is no harm in trying. Go on, do it. Chemistry is an experimental science, they say. Though I wouldn't expect much of anything. Suspensions usually settle; true solutions do not. $\endgroup$ – Ivan Neretin Dec 12 '16 at 18:15
  • $\begingroup$ Your source has a somewhat dim and unscientific view of standard domestic water that isn't justified in reality (unless you live in Flint, anyway). In western countries municipal tap water has to meet a range of standards about what it contains and usually does (this, BTW, is one of the single biggest innovations in human health in the last 200 years). This means that further filtration or treatment is usually useless. OTOH, supplies which are hard may show changes over time as minerals precipitate (they don't settle, they are in solution). But these changes won't affect health. $\endgroup$ – matt_black Apr 28 '17 at 19:38
  • $\begingroup$ @matt_black, I do not agree that water filtration is useless. For example, read about chromium-6 in drinking water. Here is a nice and concise article written by a PhD in chemistry with a lot of background in water filtration: here $\endgroup$ – N4v May 1 '17 at 17:08
  • $\begingroup$ @N4v The first rule of Skeptics club is: don't believe anything you read in marketing material. Your source is trying to sell you something. that doesn't mean what he says is wrong but it does mean you can't trust it unless you find independent verification from someone who isn't trying to sell you something. $\endgroup$ – matt_black May 1 '17 at 23:05
  • $\begingroup$ @matt_black, I understand your logic, but think of these two facts: (1) there are many substances which have been verified through independent research to be harmful. These include things like lead, copper, and disinfection byproducts. (2) The EPA set contaminant limits not based on ideal levels, but based on the balance between health and governmental practicality. Do you disagree with either of the things I just said? $\endgroup$ – N4v May 2 '17 at 23:02
3
$\begingroup$

I have experimented with this phenomenon and found that allowing water to sit for a day may have a non-trivial effect on the TDS gradient in the water. This effect could be due to various reasons, however.

Materials: Glass pitcher, plastic & porcelain & glass containers; tap water from near Chicago, IL, USA; TDS meter

Methods: I poured tap water into various containers and allowed the water to stand for about 30 hours. I then measured the TDS at the top of the water. Then I stirred the water vigorously and let the water sit for about a minute. I took a TDS reading again at the top of the water. I repeated the experiment a total of 4 times.

Results (Updated 12/23/2016): $$\begin{array}{ccc} & \text{TDS Before Stirring} & \text{TDS After Stirring} \\ \hline \text{Glass} & 138 & 141 \\ \text{Plastic} & 150 & 146 \\ \text{Porcelain} & 153 & 152 \\ \text{Glass} & 147 & 149 \\ \end{array}$$

Conclusion: The results are interesting. I will discuss a few possible reasons behind the results, but I'm not certain.

  • TDS (after) was higher both times for glass. Maybe this is because when the water sits, a small number of ions adsorb to the glass surface.

  • For plastic and porcelain, TDA (after) was lower for both plastic and porcelain. This may be due to dust that settled and dissolved into the water, as I did not cover the containers (looking back, covering the containers would probably have made the experiment more controlled). In addition, maybe ion adsorption happens less with porcelain and plastic.

$\endgroup$
  • 1
    $\begingroup$ How many times did you measure it? Also, what is the reading for the fresh tap water? $\endgroup$ – Ivan Neretin Dec 16 '16 at 18:50
  • 2
    $\begingroup$ Indeed. Per @IvanNeretin's comment, you should repeat the experiment at least three or four times and calculate mean/st.dev. on the before/after difference. $\endgroup$ – hBy2Py Dec 16 '16 at 19:21
  • 1
    $\begingroup$ Have you excluded other explanations, eg. ions absorbing on the glass wall? $\endgroup$ – Greg Dec 17 '16 at 0:55
  • $\begingroup$ Give me some time and I will repeat the experiment. Easy with the downvotes, though. No need to hate on this. This is a real experiment. @Greg, that is a good question. I will think about that $\endgroup$ – N4v Dec 17 '16 at 12:50
  • $\begingroup$ Are those differences even significant? How do you tell noise form signal here? $\endgroup$ – matt_black Apr 28 '17 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.