Chemistry, mixing different pH solutions, is this right?
Discussion
Basically I am trying to get a handle on the pH of a final goop of chemicals
I have the pH of each and I have the amounts in liters
So, in short, what I want to know is if this seemingly straight forward calculation is correct: (see answer by Trevor H)
https://answers.yahoo.com/question/index?qid=20100...
Now, there is one issue where there is a chemical of 10 L of 50 parts water, 1 part Ammonium Hydroxide (pH 3) and 1 part H2O2 (pH 8).
Am I right in saying if 1 part = 1ml, then in 52 ml I have 50 ml water, 1 ml Ammonium and 1 ml H2O2?
If so can I work out the pH following trevor H example and arrive at a pH of 4.71?
I did use this: http://www.webqc.org/phsolver.php putting in the chemical formulas, concentrations and kPa values and got a pH of 5.5. Not sure I did it right mind.
I have the pH of each and I have the amounts in liters
So, in short, what I want to know is if this seemingly straight forward calculation is correct: (see answer by Trevor H)
https://answers.yahoo.com/question/index?qid=20100...
Now, there is one issue where there is a chemical of 10 L of 50 parts water, 1 part Ammonium Hydroxide (pH 3) and 1 part H2O2 (pH 8).
Am I right in saying if 1 part = 1ml, then in 52 ml I have 50 ml water, 1 ml Ammonium and 1 ml H2O2?
If so can I work out the pH following trevor H example and arrive at a pH of 4.71?
I did use this: http://www.webqc.org/phsolver.php putting in the chemical formulas, concentrations and kPa values and got a pH of 5.5. Not sure I did it right mind.
I did some number crunching, calculating the proton concentration of the 50 vol. water with 1 vol.
NH4OH:
[H+] conc in pH3 = 0.001 M
Dilute in 50 vol. water (=51): 0.001 / 51 = 1.96 x 10^-5 M
pH = -log[H+] = 4.7
H2O2 at pH = 8 comes out as a H+ conc of 1 x 10-8, so it's addition to the overall concentration of hydrogen ions is almost negligible here (diluted up to 52 vol liquid adds 1.92 x 10^-10 M to the concentration).
Keeping with sig figs being about 2, I got a pH of your 50 vol. water / 1 vol ammonium hydroxide / 1 vol peroxide to be pH = 4.7.
can you get a pH meter to give you a ballpark measurement? You could also do some titration (get about 5 runs, neutralizing to the end point with e.g sodium hydroxide (might have to standardize a solution first!) and take an average then work it all out)
NH4OH:
[H+] conc in pH3 = 0.001 M
Dilute in 50 vol. water (=51): 0.001 / 51 = 1.96 x 10^-5 M
pH = -log[H+] = 4.7
H2O2 at pH = 8 comes out as a H+ conc of 1 x 10-8, so it's addition to the overall concentration of hydrogen ions is almost negligible here (diluted up to 52 vol liquid adds 1.92 x 10^-10 M to the concentration).
Keeping with sig figs being about 2, I got a pH of your 50 vol. water / 1 vol ammonium hydroxide / 1 vol peroxide to be pH = 4.7.
can you get a pH meter to give you a ballpark measurement? You could also do some titration (get about 5 runs, neutralizing to the end point with e.g sodium hydroxide (might have to standardize a solution first!) and take an average then work it all out)
Edited by thatdude on Wednesday 8th October 13:55
Don't think the Trevor H calc is correct. If you take his method, using say 100 ml of 1 MOL HCl (pH=0) and 100 ml of 1 MOL NaOH (pH=14) you get:
HCl solution:
[H+] = 10^0 = 1.0 M solution of H+ ions
100ml solution will contain 0.1 mol H+ions
NaOH solution
[H+] = 10^-14 M solution of H+ ions
100ml will contain 10^-15 mol H+ions
Total H+ions = 0.1 M in 200 ml solution
Molarity of H+ ions, [H+] = 0.5
pH = -log [H+]
pH = -log 0.5
pH = 0.30
Clearly this is completely wrong as in reality you get a solution of salt water with a pH of 7. The problem being he's counting up the hydrogen ions but ignoring the effect of adding additional hydroxide ions. So his method is approximately correct for mixing solutions of exclusively strong acids, or of exclusively strong bases, but not for mixed acid/base mixtures and not for weak acids or weak bases.
HCl solution:
[H+] = 10^0 = 1.0 M solution of H+ ions
100ml solution will contain 0.1 mol H+ions
NaOH solution
[H+] = 10^-14 M solution of H+ ions
100ml will contain 10^-15 mol H+ions
Total H+ions = 0.1 M in 200 ml solution
Molarity of H+ ions, [H+] = 0.5
pH = -log [H+]
pH = -log 0.5
pH = 0.30
Clearly this is completely wrong as in reality you get a solution of salt water with a pH of 7. The problem being he's counting up the hydrogen ions but ignoring the effect of adding additional hydroxide ions. So his method is approximately correct for mixing solutions of exclusively strong acids, or of exclusively strong bases, but not for mixed acid/base mixtures and not for weak acids or weak bases.
On the subject of your original question, I think something is a bit wrong as ammonium hydroxide is a base, so would have a pH > 7 (not 3!) and hydrogen peroxide is an acid so would have a pH < 7 (not 8). Did you get those numbers backwards maybe?
I took your two solutes both as 1 part in 26 solutions in water and solved using the linked solver, it gave me a pH of 10.47 (pKa/pKb values pulled off wikipedia):
Name: NH3
Initial concentration after solutions are mixed = 0.0385
Kb1 = 1.7782794100389E-5 ( pKb1 = 4.75 )
Name: H2O2
Initial concentration after solutions are mixed = 0.0385
Ka1 = 1.7782794100389E-12 ( pKa1 = 11.75 )
Answer:
[H+] = 3.4102777025924E-11
[OH-] = 0.00029323125188305
pH = 10.467210254499
pOH = 3.5327897455007
I don't think this is what you were after though, as your states pH values don't correspond to those initial concentrations.
I took your two solutes both as 1 part in 26 solutions in water and solved using the linked solver, it gave me a pH of 10.47 (pKa/pKb values pulled off wikipedia):
Name: NH3
Initial concentration after solutions are mixed = 0.0385
Kb1 = 1.7782794100389E-5 ( pKb1 = 4.75 )
Name: H2O2
Initial concentration after solutions are mixed = 0.0385
Ka1 = 1.7782794100389E-12 ( pKa1 = 11.75 )
Answer:
[H+] = 3.4102777025924E-11
[OH-] = 0.00029323125188305
pH = 10.467210254499
pOH = 3.5327897455007
I don't think this is what you were after though, as your states pH values don't correspond to those initial concentrations.
Edited by Flibble on Thursday 9th October 00:26
Thanks for the input guys. The problem isn't really mine to solve. I was trying to help some one out.
And yes I did go away and look up the individual components and their pH value s are not what I was given. However, I believe this is what the pH values are once the fluids have been used for what ever process they are needed (etching for lithography, making circuits and semiconductor ICs). From what i read the fluids get more acidic during the processes. Or they could be the wrong way round! Lol!
More information required and more research required!
And yes I did go away and look up the individual components and their pH value s are not what I was given. However, I believe this is what the pH values are once the fluids have been used for what ever process they are needed (etching for lithography, making circuits and semiconductor ICs). From what i read the fluids get more acidic during the processes. Or they could be the wrong way round! Lol!
More information required and more research required!
Edited by Otispunkmeyer on Friday 10th October 23:01
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