Hello everyone.
So an article was posted recently that caused a bit of a kerfuffle on r/NoPoo: it
showed that when raw baking soda is added to water, the pH remained more or
less the same, regardless of the amount of water added. Only when she took some of the bakingsoda
water solution (i.e. AFTER the solution is in equilibrium) was dilution possible.
I thought it might be instructive to look at why that is. Not entertaining, particularly, but instructive. There's a metric buttton of equations in here, which is why I changed from just posting on reddit to the blog.
–Let’s do some conversions and define some terms–
1 tablespoon = 0.0148 L
1 cup = 0.237 L
2 cups = 0.473 L (Not dodgy math: the values of 1 cup and 2
cups were calculated independently and rounded to three significant figures)
Baking soda =
NaHCO_{3} (Take heed! Baking soda is a chemical. It's not really even a natural one most of the time.)
–There are a few factors that complicate this reaction–
First of all, pH is the measure of acidity or basicity in aqueous solutions. That is to say, before we can talk
about what the pH of NaHCO3 “is”, we first have to make a solution with water.
Furthermore, NaHCO_{3} is a weak base, which means
it does not dissociate completely in water (Rats! This means loads more work
for us).
What happens when you mix NaHCO_{3} with water?
The Na^{+} ion is a “spectator ion”, leaving us with
HCO_{3}. This gets even trickier because HCO_{3} is amphiprotic, meaning it can act like an
acid OR a base, depending on the reaction. HCO_{3} can react to form its conjugate acid H_{2}CO_{3} (carbonic acid) and OH^{}
(what makes a solution basic), and viceversa, although the system will
eventually reach equilibrium. We
need to know the acid dissociation
constant (k_{a}) and base
dissociation constant (k_{b}).
The k_{a} of a weak acid can be used to find the k_{b}
of its conjugate base, using k_{a}*k_{b}=1x10^{14}. The k_{a} of carbonic acid is 4.45x10^{7},
so the k_{b} of HCO_{3}
is 2.25x10^{8}.
–Let’s get started!–
We have to do a few preliminary calculations.
First, we need to know what the molarity of our baking soda solution. Let’s assume that we are making the standard 1 tablespoon of
baking soda to 2 cups of water nopoo mix, since this is the most
frequentlygiven ratio, the one that people are advised to try first and then
adjust based on results.
We need to know what 1 tablespoon of baking soda is in
grams, which is a bit of a pain, because one is a measure of volume and the
other of mass. There’s no clear
conversion. If you try and find out by googling,
you’ll find kind of a range of answers.
So, I figured I’d just measure a tablespoon for myself and see what I
got. It came to 12.6 grams, which is in range, so this
is the figure I’ll be using for the calculation.
The molar mass of NaHCO_{3} is 84.007 grams so to
find how many moles in a tablespoon:
Now, we need the molarity of the solution so:
We will get one mole of HCO_{3} from NaHCO_{3},
so we can say that the preequilibrium concentration of HCO_{3} will be
the same as above. The OH^{}
concentration is so small that it can, helpfully, be regarded as 0M
(technically speaking it’s 1.0x10^{7}M). Since there is no H_{2}CO_{3} before
reaching the equilibrium state, it is also 0M.
After reacting with water, one mole of HCO_{3} will
form one mole of OH^{1} and one mole of H_{2}CO_{3}. So, given that, let’s make a table to
chart these changes:
HCO_{3}

H_{2}CO_{3}

OH^{}


PreEquilibrium

0.315 M

0

0

Change

x

+x

+x

Equilibrium

(0.315x)M

xM

xM

Now we get to use the equilibrium equation:
Now we have something we can work with! We know that the K_{b} of HCO_{3}
is 2.25x10^{8} from earlier.
It looks like here we have to use the quadratic equation to
solve, but it turns out that since x is likely to be much, much smaller than
0.315, you can drop the x from the denominator without affecting the outcome
too badly (you can doublecheck that this is a valid approximation
later onjust put it back into the equation for K_{b} above and make sure it all checks out). So:
Since “x” stands for the molarity of OH^{}, we can
now use this value to solve for the pH of our solution!
So you can see that 1
tablespoon in 2 cups of water will have a pH of 9.9. This is very high, and exactly what the other blogger
demonstrated with universal indicator paper.
You can also see why
adding more water doesn’t really do much to the pH of the solution. Adding more water will only change the
initial Mvalue that we used. Let’s
look at this equation with 20 cups (= 4.73 L) of water.
Almost no different!
So, what value of M value do you need in order to have a pH of 7? Well, let’s work backwards. If our pH is 7, so is the pOH. Then, we can use the definition of pOH
to find that we need x=10^{7} to get a neutral solution. Now our equation looks like this:
This is a tiny, tiny number! Let’s keep going to see how many liters of water you need to
add to a tablespoon of baking soda to get this result:
–TL;DR–
1 tablespoon of baking soda in 2 cups of water has a pH of 9.9.
1 tablespoon in 20 cups has a pH of 9.4.
You need 1418439 cups of water to make a solution with a pH close to 7.
Anyway, the key point of all of this is, because of the nature of the reaction that NaHCO_{3} has with H_{2}0, it takes an enormous amount of water to dilute even one tablespoon of baking soda down to a nearneutral pH.
1 tablespoon in 20 cups has a pH of 9.4.
You need 1418439 cups of water to make a solution with a pH close to 7.
Anyway, the key point of all of this is, because of the nature of the reaction that NaHCO_{3} has with H_{2}0, it takes an enormous amount of water to dilute even one tablespoon of baking soda down to a nearneutral pH.
NoPoo People who use baking soda (or those who are
considering it) may want to take this fact into consideration when constructing
their hair care regime. It’s
certainly true that pH is not THE ONLY important factor when it comes to hair
health, but it’s also a fact that the scalp has an acid mantle that will be
disrupted by having a highpH solution onto it, and there are a number of
people who say that hair itself is strongest when slightly acidic.
Furthermore, the pH of 1 tablespoon : 2 cups solution is
starting to approach 10, which is the range where the cuticle of hair can becaused to lift. This is very
undesirable. As the linked page
says, the hair cuticle is not hinged, it cannot open and close freely! Repeatedly lifting and smoothing it
will damage it.
Those of us striving for a scientificallysound hair care
regime would do well to leave the baking soda in the kitchen.
If you'd like to know more about alternatives to baking soda, I'll be testing out common choices for a month at a time. So far: rye flour.
If you'd like to know more about alternatives to baking soda, I'll be testing out common choices for a month at a time. So far: rye flour.
Thank you for the awesome chemistry explanation! I was looking for something like this :)
ReplyDeleteI'm glad it was interesting! Thanks for reading (ˆˆ)
DeleteHello, thank you for taking the time and effort to make all these calculations that were triggered by my findings on baking soda: http://blog.kanelstrand.com/2014/01/bakingsodadestroyedmyhair.html I was wondering if you would allow me to repost this post as an update to my original article, with a link back to your blog of course.
ReplyDeleteI wonder if you can do equations like this for soap..
ReplyDeleteI'd actually really love to post the equations for soap as well, but they're a little bit more complicated. Soaps are made of salts of fatty acids, and there can be different types and compositions, whereas baking soda is pretty simple. I've also been kind of distracted with my Game of Thrones costume analysis project, but I'd like to find time to work out a (simplified) version of the soap calculations as well. I can say based on testing with Universal Indicator paper, though, that the pH of soap also doesn't dilute much.
Delete