What is the pH of...

[chiddler]

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What is the pH of a solution where the hydroxide concentration is 10^6 times greater than hydronium concentration?

A. 10.

I don't feel that I really grasp the essence of this question though. My first mistake (as TBR predicted) was adding 6 to neutral pH, 7, which is 13. They explained that this is improper usage of log scale.

Anybody mind providing an explanation why this is improper log scale usage?

Thank yous.

 

[milski]

---- wrong - do not use, see post #8

log[OH-]-log[H+]=6 // from the problem
log[OH-]+log[H+]=14 // generally true for water in normal conditions

log[OH-]=10
---- wrong - do not use, see post #8

Note that the hydroxide concentration is 10^6 the concentration of the hydronium, not 10^6 the concentration of hydroxide in neutral solution. By doing 7+6 you are doing the latter.

 

[chiddler]

I'm doing something wrong:

log[OH] = 6 + log[H]
log[OH]+log[H]=14

6+log[H]+log[H]=14
2log[H]=8
log[H]=4
pH = 4

Note that the hydroxide concentration is 10^6 the concentration of the hydronium, not 10^6 the concentration of hydroxide in neutral solution. By doing 7+6 you are doing the latter.

Can you please explain this. pH = 7. If pOH is 6 more, then pH drops by 6 to pH = 1, and pOH =13. Why is this wrong 🙁

 

[milski]

I'm doing something wrong:

log[OH] = 6 + log[H]
log[OH]+log[H]=14

6+log[H]+log[H]=14
2log[H]=8
log[H]=4
pH = 4

---- wrong - do not use, see post #8
Line 3 above is wrong. If you add the two equations:
2log[OH]=8
log[OH]=4
log[H]=14-4=10
You can plug in the original two equations to verify that log[H]=10, log[OH]=4 is a solution.
---- wrong - do not use, see post #8

Can you please explain this. pH = 7. If pOH is 6 more, then pH drops by 6 to pH = 1, and pOH =13. Why is this wrong 🙁

pOH is not 6 more. pOH is 6 more than pH.

Let's say that you always have 14 fruits.
First you had 7 apples and 7 oranges. Then you had 6 more apples. That would make 13 apples and 1 orange, since applesNew = applesOld + 6. That's your solution/interpretation.
But the problem says that your apples are six more than your oranges, as in apples = oranges + 6.

Does it make more sense now? If not, I'll have to come up with some other way to explain it.

 

[SaintJude]

I still don't get it. chiddler, can you please post the book's solution ?

 

[chiddler]

Ah I understand the analogy. But now i'm trying to figure out why my math is not working out. I understand adding the two equations together; that works out perfectly. But why isn't my algebra working! Why is line 3 incorrect mathematically?



Book solution:
Because hydroxide is greater in concentration than hydronium, the solution is basic. The trick here is to use the log scale correctly. Because the concentrations differ by a factor of 10^6, the pH and pOH differ by 6. If the pH is 13, then the pOH is 1 and the difference is 12. This is a tricky question where understanding the conversion between the log scale and concentration is pertinent. Most students pick 13...because...yadda yadda.

Not the most helpful solution.

 

[milski]

Never mind, I screwed up, my math is wrong - 6 + 14 is 20, not 8. pH is negative log and that will change the first equation - replace 6 there with -6. After that it should work out ok.

Yes, the explanation sucks. It's not a problem with understanding the log scale but not taking into account that the pOH also changes and needs to be treated as variable instead of using the constant 7.


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[milski]

Ok, I made a mistake in my post last night.
log[OH]-log[H]=6 // still correct, from the problem
log[OH]+log[H]=-14 // pH is negative log!

Solvig this gives us log[H]=-10, which means that pH is 10

Or if you want to work directly with pH/pOH:
pOH+pH=14
pH-pOH=6 // same as -pH - (-pOH)=6

Which gives us pH=10

 

[chiddler]

Never mind, I screwed up, my math is wrong - 6 + 14 is 20, not 8. pH is negative log and that will change the first equation - replace 6 there with -6. After that it should work out ok.

Yes, the explanation sucks. It's not a problem with understanding the log scale but not taking into account that the pOH also changes and needs to be treated as variable instead of using the constant 7.

Ok so it's:

log[OH] = -6 + log[H]
log[OH]+log[H]=14

Now the numbers make sense, but I don't understand this conceptually. So how would I know to use -6?

 

[milski]

Ok so it's:

log[OH] = -6 + log[H]
log[OH]+log[H]=14

Now the numbers make sense, but I don't understand this conceptually. So how would I know to use -6?

Sorry, I posted that one too quick and instead of clearing up things made them worse.

If you write the equations with log[X], you have to change the second equation, the sum of the logs is -14.

If you're writing them for the pH, you have to change the signs, since pX=-logX, so log[OH]-log[H] = 6 becomes -pOH - (-pH)=6 or -pOH+pH=6 or pH-pOH=6.

 

[chiddler]

Sorry, I posted that one too quick and instead of clearing up things made them worse.

If you write the equations with log[X], you have to change the second equation, the sum of the logs is -14.

If you're writing them for the pH, you have to change the signs, since pX=-logX, so log[OH]-log[H] = 6 becomes -pOH - (-pH)=6 or -pOH+pH=6 or pH-pOH=6.

Ok!! This makes much more sense conceptually too!

thanks very much

 

[FishyTheFish]

Simple way of doing this.
you know that log[H+] + log[OH-] = 14
Using log rules, this is the same as

[H+][OH-] = 10^-14
Let x = [H+] and we know that [OH-] = 10^6*[H+] or [OH-] = 10^6*x
so
x*x*10^6 = 10^-14
x^2 = 10^-20
x = 10^-10

pH = -log[H+] = 10

 

[SaintJude]

Simple way of doing this.
you know that log[H+] + log[OH-] = 14
Using log rules, this is the same as

[H+][OH-] = 10^-14
Let x = [H+] and we know that [OH-] = 10^6*[H+] or [OH-] = 10^6*x
so
x*x*10^6 = 10^-14
x^2 = 10^-20
x = 10^-10

pH = -log[H+] = 10

Thank you! This is much more intuitive to me. This way, I understand that the only equation you needed to remember was that [H+][OH-]= 10^-14.