Originally Posted by blueberrydrumz
basically if i wouldnt break the buffer of my tap it would slowly rise towards 7 ... so i was asking if using a lot of P2O5 to take the water down to 4.5 to break the buffer and then bringing it back up to 5.5 will unbalance the water and maybe in the end unbalance my feed water?
Or will that be the case anyway as we want a acidic solution of around 5.5... and it doesnt matter how i get to 5.5 as H+ and OH- will never be in equlibrium in the feeding water as we are shooting for 5.5?
hope im getting my point across hehe
K maybe I got your point... Not so easy to explain but... Kinda interesting...
pH=-log[H+] (for example pH7=10-7 H+)
while pOH=-log[OH-] (so pOH7=10-7 OH-)
also, pH+pOH=14 so if pH=5.5 than pOH=8.5 no matter how you get there.
So, if by "unbalanced" you mean different concentration of H+ and OH- yes, otherwise you would be at pH7.
BUT, the 2 path are different in terms of substances added to water so of course there must be some difference when when you finally reach pH5.5
It's related to buffer, I'll try to explain.
so, P2O5 reacts with water giving H3PO4 wich is a strong acid. It has 3 dissociation levels (it can release up to 3 H+ ) so I'll use HCl instead just to make it easier.
Also, water has many ions acting as buffers so I'll use a classic "chemistry classroom" example of acetic acid in balance with his ion.
Basically a buffer is just something that can "neutralize" both OH- and H+ that are eventually added to a solution. So lets suppose we have a solution with acetic acid (CH3COOH) and its ion (CH3COO) at the same concentration of 0,1M
CH3COOH + H2O = H3O+ + CH3COO-
To calculate corresponding pH we use that formula:
because we said they are at the same concentration of 0,1M, H+ concentration will be equal to Ka (dissotiatiom factor of the acid) and pH will be equal to pKa, in that case 4,74.
If we add 0,01M of HCl, it will react with acetate
CH3COO- + HCl → CH3COOH + Cl-
Note that HCl has dissociated in H+Cl- but H+ bonded with CH3COO- so H+ concentration didnt raise because of them.
At the same time, 0,01M of CH3COO- reacted with HCl so CH3COO- concentration will be 0,10-0,01=0,09.
Knowing new concentrations and Ka we can calculate new pH with the same formula as before
in that case will be 4,66 just 0,08 lower than before. Consider that adding the same amount of HCl to pure water would bring pH from 7 to 2 as 0,01M HCl would mean0,01M Cl- and 0,01M H+ . 0,01 = 10-2 so pH 2
Obviously, everything said above is true (at his countrary) in casr a base is added
CH3COOH + NaOH → CH3COO- + Na+ + H2O
If you keep adding HCl (or NaOH) it will react with CH3COO- (or CH3COOH) till there will be no more left.
So, going back to your question, you will always end with the same pH of 5.5 so same concentration of H+ and OH-, but you will have a lot more PO4--- (and other dissiociation forms too) along with the ion of the base you use to raise it from 4.5 to 5.5
I know its kinda complicated but I cant explaine it better than that