[orange gold]

500mL of 2M NaOH + 200mL H2O... What is the final pH?
I got approx. 12.07...
Is this the right answer?
How do you solve it?

[XHNO20]

500ML 2M NaOH= 1 Mole of NaOH present. (2M/1L  *.5l)

200ML+500ML=700ML total solution

1 mole NaOH/700ml=Molarity=.00142M NaOH

-log(.00142)= 6.55 pOH

pH+pOH=14

14-6.55=pH

pH=7.45

[rabolisk]

1 mole NaOH/700ml=Molarity=.00142M NaOH

Error. Doesn't it seem weird that adding 200 mL to a 500 mL solution would reduce the concentration of NaOH from 2M all the way to .00142M? That is like dilution by a factor of over a 1000. I wonder if it has something to do with units...   

[Borek]

-log(.00142)= 6.55 pOH

No. Do you know the difference between natural log and base ten log?

Funny thing is that apart from many mistakes on your way to the result, your general idea is correct.

[orange gold]

I still don't know what I'm doing wrong though? I tried getting the molarity of OH.. (1 moles / 0.7L)
and to find the pOH = -log( (1/0.7) ) = -0.155
pH = 14- (-0.155) = 14.155???
This doesn't seem correct at all

[Nobby]

Calculation is wrong

2 M NaOH means 2 mol/l.  Wie have 0.5 liter, what means we have 1 mol NaOH.

This will be diluted to 700 ml what means 1 mol/0.7l = 1,428 mol/l

pOH = -log(1,428) = 0,15 => pH = 14,15

This happens , because Molarity is higher then 1.

[orange gold]

*Negative log
-log(1,428) = -0,15

[orange gold]

The reason I got this answer originally: "12.07" was because I had 14.155 but thought it was crazy that it was greater than 14.0 so I took the average of 14.155 and 7.0 (7.0 coming from the H2O) Don't know why I did this, I was just trying to get a number back on the scale of 0-14...

So if M is greater than 1 you get insane pH values that are literally off the 0-14 scale?
But overall I did the conceptual problem right (minus that last "averaging" part at the end)? It's just an outrageous problem?
If I solved the same problem but said 0.1M instead of 2M NaOH would I get a more reasonable answer?

[rabolisk]

Even if you averaged those values, you wouldn't get 12.07.

There is nothing outrageous or crazy about this problem. By definition (not correcting for activities), pH = -log[H⁺], and mathematically, proton concentration greater than 1M will give you a negative pH. By the same reasoning, hydroxide concentration greater than 1M will give you >14 pH. Of course, you probably will not be able to easily measure pH in this range with your electrode, especially considering that the calculated pH and actual pH will be different due to ionic strength, and that electrodes are usually calibrated to ranges closer to 7 (typically 4, 7, and 10). But as a theoretical calculation problem, there is nothing wrong with this problem or the answer. Remember, math never lies, if it is applied correctly.

[orange gold]

So what your saying is I will pass my test today??
Thanks for the help forum!

[AWK]

Since activity coefficient for ~1.5 M NaOH is  ~0.7 real pH will be lower and close to 14.

[aeacfm]

There is nothing outrageous or crazy about this problem. By definition (not correcting for activities), pH = -log[H⁺], and mathematically, proton concentration greater than 1M will give you a negative pH. By the same reasoning, hydroxide concentration greater than 1M will give you >14 pH. Of course, you probably will not be able to easily measure pH in this range with your electrode, especially considering that the calculated pH and actual pH will be different due to ionic strength, and that electrodes are usually calibrated to ranges closer to 7 (typically 4, 7, and 10). But as a theoretical calculation problem, there is nothing wrong with this problem or the answer. Remember, math never lies, if it is applied correctly.

great , but if i want to report the result what can i say ?
would it be pH = 14 in case of base or pH >14
also pH = zero in acids or pH <Zero

[rabolisk]

If you are answering a question like this on paper, you report what you get from math. If you measure it in lab, then you report what you get through your instrument. My point was that what is easily observable in theory isn't always so in practice. It's kind of like stoichiometry. You will never get 100% yield in practice, but you can calculate 100% yield in theory.

[aeacfm]

If you are answering a question like this on paper, you report what you get from math. If you measure it in lab, then you report what you get through your instrument. My point was that what is easily observable in theory isn't always so in practice. It's kind of like stoichiometry. You will never get 100% yield in practice, but you can calculate 100% yield in theory.

i can get your point but what made me saying that pH meter (hach instrument) gave results for  pH of unknown to be 16 some times 15.5 and also in minus some times at that moment i didnt know how to report that so i report it as 14 and zero , and here i find the chance to ask
any way thank you for the reply

[Nobby]

i can get your point but what made me saying that pH meter (hach instrument) gave results for  pH of unknown to be 16 some times 15.5 and also in minus some times at that moment i didnt know how to report that so i report it as 14 and zero , and here i find the chance to ask

In a report you write down your results you observed, after confirmation and verifying. If you find pH 15 then you write this down.