pH and Molarity

Name: Jose
Status: student
Grade: 9-12
Location: N/A

Question: Dear Argonne scientists, I am a 11th grade student in
Peru and i will carry out an experiment showing how P.H of water
used for watering certain bean plans would affect their growth
(height, biomass etc). The problem is, we can't use pH papers,
litmus papers, so we will have to measure its acidity or how
alkaline it is by Molarity. If i want to chose 6, 6.5, 7, 7.5, 8 as
my range of pH's, then how can i calculate the Molarity of each?
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Jose Luis,
 
You may find this process a little bit difficult in terms of precision 
if you are using a strong acid. The problem is that you are shooting 
for pH's that are very nearly neutral (pH = 7) and so you would have 
to obtain very dilute strong acids in order to get a pH in the range 
of 6 - 6.5.
 
The equation is from the definition of pH which is:
 
pH = -log([H+]) where [H+] = M of the strong acid
 
and converting this to solve for the target molarity you would use:
 
1 / 10^(pH) = M
 
and as you can see, to get a pH of 6.5 you would need a M that is 
3.16 x 10^-7. And as you can see, this is very dilute and very hard 
to obtain.
 
On the other hand, if you use a weak acid then it's still pH = -log([H+]), 
but the [H+] this time is not just the molarity of the weak acid. 
Rather:
 
[H+] = Square root of (ka x M of the weak acid)
 
Thus, if you wanted to get a pH = 6.5, you would need an 
[H+] = 3.16x10^-7 which is equivalent to:
 
([H+])^2/ka = M
 
and depending on the ka of your acid, you would still be working 
with very dilute solutions. For example if you use acetic acid which 
has a ka = 1.8 x 10^-5 then you would need a M that is 5.6 x 10^-9.
 
For the range you are looking for, you might be better off using 
solutions of weak acid and weak base SALTS or buffer solutions. You 
can find tables of typical salt solutions or buffer solutions to 
prepare particular pHs, rather than to dilute acids.
 
Greg (Roberto Gregorius)
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