Finding logarithm of a non whole number

[premed56]

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Ok so I've been using BR books, and I have come across a solution to a question that they don't clearly explain.

Q) what is the pH of .00391 MKOH

Their soln in order to eliminate wrong choices is to find the possible range that the pH could be like this:

3-log 10< 3-log 3.91 < 3-log 3. Which would equal 2< 3-log 3.91 < 2.5


Why do they use log of 10 and 3??

Could someone please explain? Thanks!

 

[Odi]

I don't like how TBR treats logarithms. Kaplan has an amazing method explained in their book in the acids and bases section that literally takes me 10-15 seconds max to approximate the log of any number.

Turn your number into scientific notation quickly so 3.91 x 10^-3

Then the log of this number is always [the power - log (number being raised to that power)]

In this case [3 - log(3.91)]

At this point you just have to be able to estimate the log of 1 to 10. Log of 3 = 0.47 which I remember as approx 0.5... so answer is about about 2.5.

EDIT: Subtract this number from 14 to get pH as Cmdr pointed out so pH is around 11.5

Also if the power on 10 is positive... than you just add instead of subtract.

It's by far the best method I've found. Practice it with a bunch of numbers to get the hang of it.

 

[Cmdr_Shepard]

I normally just convert it to scientific notation first. So 3.91 x 10^-3. But this is OH- concentration.

Then the pOH is between 2 and 3. Around 3.15 is where the exponent changes by 0.5 points, so you know you will subtract just over 0.5 from 3, because a pOH of 3 is equal to an OH conc. of 1 x 10^-3, and 3.91 x 10^-3 is more than this.

Therefore the real pOH is somewhere slightly below 2.5. Then you do the subtraction from 14 to convert to pH which gives 11.6 or so. On the real MCAT, there will only be one answer close to a pH of 11.6.

 

[siffster]

0.0039 M = 3.91*10^-3 M = ~ 4*10^-3 M.

take the -log --> -log [4*10^-3] ---> -log [4] + -log [10^-3] ---> -log[2*2] + -log[10^-3] ---> -log 2 + -log 2 + -log[10^-3]. since the log of 2 is 0.3, -log 2 = -0.3. so you get: -0.3 + -0.3 + -log[10^-3] = -0.6 + 3 = 2.4

if you just remember that the log of 2 is 0.3(although I've had that ingrained in my brain since my college years), and that the log 0f 3 = 0.5, you can pretty manipulate any number using the log rules, which is what TBR teaches you