calculating pH / log rules

[DrMattOglesby]

alright, so pH = -log[H+]

what is the pH when: [H+] = 7 x 10^-4 ?

can the log be rewritten as: pH = -(log7 + log10^-4) ?

if i do that, I get log7 approximately = to .9 and log10^-4 = to -4 inside the parentheses.
then i distribute the negative sign and end up with pH = 3

did I break any rules??? Does a pH of 3 "sound" sensible with that [H+] ????

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[tncekm]

For the MCAT the pH= 3 would be incorrect rounding.

pH (7 x 10⁻&#8308😉 is "less than 4 and more than 3" because:
* Log10 = 1
* Log1=0
* Log(1)=0 < Log(7) < Log(10)=1; &#8756; log 7 is between 0 and 1
* -Log(7) + -Log(10&#8315;&#8308😉 = 0>-Log(7)>-1 + [-(-4)] = 4 - Log(7)= slightly greater than 3.

In other words, your work is totally correct, and the pH would be exactly 3.15, but saying its 3 would be wrong for the MCAT and they may test you specifically on that! Its between 3 and 4, you don't need to know the specific number because there will only be one answer between 3 and 4.

 

[Vihsadas]

tncekm's math is dead on. And he's mostly correct in saying that you'll never really need to get the exact number for pH. BUT, the MCAT can test you on pHs down to about ~.5 a pH! So you need to know how to differentiate whether something is above or below of pH of say 2.5. (or 3.5 or 4.5 etc...)
Let me help you out with a neat little estimation trick.
So we know that:

[H+] of 1x10^-4 = pH 4
and
[H+] of 1x10^-3 = pH 3

But what if the MCAT asks you to differentiate between two answer choices that are something like:
a) pH of 3.15
b) pH of 3.55

So how do you deal with those 'middle' pHs? Well in the above example, remember that a multiplier of '3' in front of the x10^-4 in the H+ concentration puts you right in between two integer values of pH. Example:

Since,
[H+] of 1x10^-4 = pH 4
and,
[H+] of 1x10^-3 = pH 3
Then,
[H+] of 3x10^-4 is ~ pH 3.5

So if the multiplier is above or below '3', you can estimate whether the answer is above or below pH 3.5.

In your problem you have an [H+] of 7x10^-4, then since 7 is greater than 3, it means that the pH is LOWER than 3.5 but greater than 3! (Remember that higher H+ means lower pH.)
See how you could have done that problem in literally 2 seconds if you get comfortable with the rounding trick? You would never have to break up the log into parts and then do all that hairy math. 2 seconds is all it takes, and no paper work! 🙂

So just remember that:
[H+] of 3x10^-10 will be pH 9.5
[H+] of 3x10^-9 will be pH 8.5
[H+] of 3x10^-8 will be pH 7.5
etc...

And then if you get a weird [H+] like 8.342x10^-7, you know immediately that since 8.342 is greater than 3, the pH is going to be less than 6.5 but greater than 6...in this case since 8.342x10^-7 is closer to 1x10^-6, the pH is going to be very near to 6. (It's actually 6.078)

This little trick is sufficient for estimating these problems on the MCAT and cut your calculations down to zero! It'll save you a lot of time on these types of problems.
Did you guys understand that?

 

[DrMattOglesby]

i like that trick...
but to be honest, i just memorized these log values:

log1 = 0
log2 = .3
log4 = .6
log8 = .9
log10 = 1

I got those from MCATpearls.com (and rechecked em on my calculator).
the saying is something like, "2,4,8 equals 3,6,9"

-matt

 

[Vihsadas]

i like that trick...
but to be honest, i just memorized these log values:

log1 = 0
log2 = .3
log4 = .6
log8 = .9
log10 = 1

I got those from MCATpearls.com (and rechecked em on my calculator).
the saying is something like, "2,4,8 equals 3,6,9"

-matt

That's a cool trick to. Whatever works for you is what you should use!

 

[JLeBling]

From: http://curiousmath.com/index.php?name=News&file=article&sid=32

I came up with this method for determing the logarithm (base 10) of a number in my head back in 1995. I was tutoring a friend for her MCAT exams and a question involving pH and pKa values struck my interest. Although we were able to arrive at a suitable answer without actually needing a calculation, I still wanted a quick way of determining logarithms in the event that I found myself without a calculator (and just as a mental challenge for myself).
Here is how I do it:

1. When someone gives you any positive number, you should immediately 'write' that number in scientific notation in your head.

2. Next, focus only on the exponent of the number (written in scientific notation). This number will be the basis of you answer.

3. Estimate the logarithm of the abscissa in your head (thats the number between 1 and 9.999999..., not part of the exponent). Note: you'll need to memorize the table below (its not that hard).

4. Add the logarithm of the abscissa to the exponent you found in step 2.

What follows are the values for the logs you'll need to have memorized for step 3...

log[1]= 0
log[2]= .30
log[3]= .48
log[4]= .60
log[5]= .70
log[6]= .78
log[7]= .85
log[8]= .90
log[9]= .95

As an example, lets find the logarithm of 29,012. Written in scientific notation, that would be 2.9012 X 10^4. So, the exponent is 4. Now, we need to concentrate on the abscissa (2.9012 is very very close to 3). From our table above (which we have memorized for the trick), the logarithm of 3 is 0.48. So, we add the exponent (4), to the log of the abscissa (0.48), to get a value of 4.48. A calculator reveals how good this method is (4.46 out to two decimal places).

This method works because scientific notation is a base 10 system for writing out numbers. With some practice, you'll get a feel for how to massage your guesses for numbers that aren't exactly in the table (for instance: 2.5).

 

[adrakdavra]

Does the MCAT ask Questions that close in numbers????