[Junk Sciences] Debunking the alkaline water and pH-changing food fad with high school basic chemistry

It seems the latest fad that pops up on Facebook recently seems the whole “alkaline water” fad and its cousin “pH changing food diet” that claims it will change your body pH.

I even seen some “alkalinizing” food processor selling at a promotional price of $599 like this one:



For a biologist or healthcare practitioner, such claim makes you laugh or makes you jump off your chair. Why? Because our body is way more of a water barrel that will have its pH fluctuating that easily. Our body pH, as defined by blood and resting tissue pH is around 7.4. Blood pH is an important parameter in healthcare and a slight acidosis (pH<7.4) or alkalosis (pH>7.4) have severe consequences. There are of course a series of physiological mechanisms and a complex systems of buffers that are here to compensate and re-equilibrate the pH.

But you may still be skeptic on my claims and think that such thing may have scientific relevance. I call that scientific relevance quack.

To support my claim and debunk such fallacious fad, we will go back to high school chemistry class and basic maths. But first, let’s do the science behind.

Firstly, let’s define what is the pH. The pH is a value defining the relative acidity of an aqueous solution in a defined temperature and pressure setting. It is driven by the relative concentration of protons (H+) inside this solution.

In solution, H+ transiently exists as themselves and rapidly undergo a chemical reaction with water molecules around (because an aqeous solution is by definition full of water molecules) through the following reaction:

H+ + H2O -> H3O+ (hydronium ions).

The pH is calculated by the following formula: pH= -log [H3O+] with [H3O+] representing a molar concentration (mol/L or M). We can also obtain [H3O+] as [H3O+]=10-pH

One mole is defined according to Avogadro by the number of carbon atoms present in 12g of pure carbon. This number is: 6.22 x 1023 atoms (that’s a bit more than one billion of trillion atoms). So one mole of an element is equal to 6.22 x 1023 atoms. Carbon element (C) has a molecular weight of 12 g.mol-1, so 12g of pure carbon will contain one mole, in other means 6.22 x 10e22 atoms of pure carbon elements.

Let’s get back to our pH. As we now pH range from 1 (most acid) to 14 (most basic) for aqueous solutions. In other means, [H3O+] ranging from 10-1mol/L to 10-14 mol/L. That’s roughly a trillion mole difference.

Water is indeed a transitory form as made by by an equilibrium between H3O+ (giving the acidity) and OH (giving the basicity) as the following

H2O <-> H3O+ + OH

Pure water has a pH=7, that means [H3O+] and [OH] are in equal concentrations of 10-7mol/L.

That’s bring another concept, the concept of equilibrium constant Kw:

Kw=[H3O+][OH]=10-7*10-7=10-14 mol-2 L-2 (at 25ºC/1atm)

This constant never changes, the only parameters that changes are more you will introduce acid H3O+ in it, less OH ions will be in solution.

Now let’s get to the core of this topic: alkaline waters.

For this demonstration, I will stick to one common pH used in such water pH=8.8.

That’s give us a [H3O+]=10-8 mol/L. For reference, tap water ranges from pH 6.5 to pH 8.5. Let’s take the most acidic tap water (pH 6.5) for the demonstration.

The institute of medicine recommends 3.7L of water per day (http://iom.nationalacademies.org/reports/2004/dietary-reference-intakes-water-potassium-sodium-chloride-and-sulfate.aspx). But for this demonstration, let’s round it to 3.5L/day.

Having to drink such volume once is physically impossible and if you tried to push enough to will induce a massive electrolyte and pH imbalance that will harm your body and put you at risk of dying from hyper hydration. Thus let’s assume you will drink 1.5L of water three times a day.

Your stomach juice has an average pH=1, that makes a [H3O+]=10-1 mol/L. According to Silverthorn’s “Human Physiology” second edition textbook, it is estimated that your stomach secretes 1 to 3 L gastric juice/day. Let’s assume that your passive gastric volume is 1L.

Let’s drink 1.5L of alkaline water (pH 8.8) at once and see what’s happen to our stomach. First let’s consider the change in volume.

Our stomach (1L) contains: 1(L) *10-1(mol/L) = 10-1 mol of H3O+. By reciprocity, the amount of OH is equal to [OH]=10-14/10-1=10-13 mol/L of OH.

To summarize, our stomach contains 10-1 mol of H3O+ and 10-13mol of OH.          

Our bottle (1.5L) of alkaline water has a [H3O+] of 10-8.8 mol/L, so we have 1.5*10-8.8 mol of H3O+ a concentration of [OH]=10-14/10-8.8=10-5.2 mol of OH-.

To summarize, our alkaline water bottle contains 1.5*10-8.8 mol of H3O+ and 1.5*10-5.2 mol of OH.

By analogy the same volume of the most acidic water (pH 6.5) would contain 1.5*10-6.8 H3O+ and 1.5*10-7.2 mol of OH.

If we drink the whole water volume, the total volume now in our stomach would be: 1.5+1=2.5L.

If we drink the alkalinized water, the total H3O+ and OH amount now in your stomach would be now (1*10-1+1.5*10-8.8=1’000’000’000*10-9+1.5*10-8.8)=10-1 mol and (1*10-13+1.5*10-5.2=1*10-13+15’000’000*10-13.2)=1.5*10-5.2 mol respectively. H3O+ remains the ion present in the highest majority, almost 10’000 times higher amount than OH.

If we consumed entirely the OHin the reaction, the limiting reagent would be it:

H3O+ + OH -> H2O,

the amount of H3O+ ions remaining would be:

10-1 – 1.5*10-5.2 = 0.10000 – 0.00003 = 0.09997 mol.

The final [H3O+] would be 0.09997/2.5 = 0.039988 mol/L and the resulting pH in your stomach pH is now = -log 0.039988 = 1.398.

If you drank the same volume of the most acidic tap water, the amount of H3O+ and OH- would be (1.5*10-6.8 + 10-1 = 1.5*10-6.8 + 1’000’000*10-7)=10-1 mol and (1.5*10-7.2 + 10-13 = 1’500’000*10-13.2 + 10-13)=1.5*10-7.2 mol respectively.

the amount of H3O+ ions remaining would be:

10-1 – 1.5*10-7.2 = 0.10000 – 0.00000003 = 0.0999997 mol.

The final [H3O+] would be 0.0999997/2.5 = 0.04 mol/L and the resulting pH in your stomach pH is now = -log 0.04 = 1.397.

Before drinking water, the pH of your stomach is 1. After drinking alkaline water, the final pH is 1.398 (your stomach is 39.8% less acid). After drinking the most acidic tap water, the final pH is 1.397 (your stomach is 39.7% less acid).

Difference between alkaline and acidic pH? Less than 0.1%!

Conclusion? Alkaline water is as effective to change your stomach pH than the most acidic tap water. Worried about the 40% decrease? You should not, the body will accommodate it very well the decrease by a system of buffer and your gastric cells will automatically increase the production of protons to re-adjust the pH.

In conclusion, if you fell inside the alkaline water marketing ploy, you should get back to your high school chemistry textbook and work on your acid-base chemistry classes.