MCAT Math Vid 9 – Antilogs in pH and pKa Without A Calculator

MCAT Math Vid 9 – Antilogs in pH and pKa Without A Calculator

Leah here from and in this
video, I’ll show you how to solve MCAT style Antilogs questions without a calculator. This
video picks up from my last video where I show you how to solve logarithm based questions
and you can find this video alongwith my entire series on solving MCAT Math without a calculator
by visiting my website In the last video I showed you a trick on
how to find a pH, pOH or pKa value when given a concentration or ka. The trick showed you
that when you have a number times ten to a negative power, that power becomes your pka,
your pH or your poH. But what if now you are faced with a question
where the actual pH, poH or pka value is given and you’re ask to find the concentration or
the ka? For example you maybe given a question that says:
Find the ka of an acid whose buffer has a pH of 4.19 in a solution containing equal
moles of acid and conjugate base. I’ll cover the Science portion of this question
in my Chemistry videos at but for this video let focus just on the Math.
Since we’re dealing with a buffer we’ll use the Henderson–Hasselbalch equation which
says that pH is equal to pka plus the log of Conjugate base over acid.
You’ll also see this written as A minus over HA. Even though we have equal moles of acid
and conjugate base or fifty-fifty, whatever number we have for conjugate base is the number
for acid and that means we have a ratio of some number over itself or one.
The log of one is zero and that means this entire portion of the equation drops out telling
me that the pH is equal to the pka. Knowing that the pH is equal to 4.19 equals the pka
we know the pka is also equals to 4.19. But how do we use this to find the ka value of
this acid? 4.19 is not a clean and easy number to calculate so let’s break it down:
The first thing you want to do is check how close your answer choices are to each other
to see how much you can simplify and how quickly you can come up with the answer. Here’s the
equation we’ll use. If pka is equal to negative log of ka (pka=-log ka), since log stands
for log base ten, to solve for ka we have to have ten to the power of negative log to
cancel out and that means we need ten to the power of negative pka. So the ka value is
equal to ten to the minus pka which is equal to ten to the minus four point one nine (ka
=10^pka=10^-4.19). A nice and clean number like four would give
us a ka value of one times ten to the minus four. But we also have to account for that
4.19 so we don’t know we’re looking for the number close to one times ten to the minus
four. If this is not enough to isolate your answer, you then want to find the range where
your ka will fall out. We’ll take the number 4.19 and round it down
to 4 and up to 5. A pka of 4 has a ka of one times ten to the minus four, a pka of 5 has
a ka of one times ten to the minus five. That means the number we’re looking for is somewhere
in this range. But if this is still not enough, then you wanna go back to the trick where
I showed you how to recognize the different numbers that give you different ranges. In
review, if we have a number point one we get an eight time ten to the minus x. And I put
x instead of the number because if we have 4.19 our exponent will be a number times ten
to the minus five. So if we had 4.1 it will be eight times ten to the minus five.
A number point three will be five times ten to the minus that power. In this case, if
we have 4.19, let’s round that to 4.2, remember on the MCAT you are allowed to round because
it will be close enough. If point one gives me an eight and point five gives me a five
then our answer has to be somewhere between eight and five so all you have to recognize
is that the number is somewhere between five and eight. So it’ll be five times ten to the
minus five to eight times ten to the minus five.
Even if we haven’t narrowed in on a specific number, for the MCAT this is close enough.
In fact, punching ten to the negative four point one nine in the calculator I get an
answer of 6.46 times ten to the minus five which on the MCAT is close enough. If you
wanna narrow this down a little more, 4.19 is closer to 4.1 than it is to 4.3 and that
means we’ll be closer to the 8 than to the 5 as is evident by 6.46.
This concludes my video series on MCAT Math Without a Calculator. You can find this entire
series on my website at You can find additional MCAT videos including
Physics, Chemistry, Biology and Organic Chemistry on my website
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40 thoughts to “MCAT Math Vid 9 – Antilogs in pH and pKa Without A Calculator”

  1. Great video. May I recommend, for continuity with the previous video (MCAT Math Vid 8,) that at 3:40 you flip the table? i.e. in the previous video, the "#.1" was on the bottom, though in this video, it is on the top. That made this video a bit more challenging to follow (since I just watched the previous before this.) Thank you again for all of your help.

  2. Your videos have been very helpful, thank you! Can you help me understand how you decided that the x exponent would equal 5 at 3:53.  Please advise, thanks!!

  3. @Leah4sciMCAT Hi Leah! I love your videos! I was a bit confused in this video around 4:04. You didn't mention how you determined that the exponent is going to be -5 and not -4 or some other number. Thanks!

  4. I realize this question has been asked before, but I don't think you gave a satisfactory answer. Since our desired number has the exponent lying between 4 and 5, why did you pick 5 instead of 4 as the exponent? In other words, why didn't you pick our value ranges to be between: 5 x 10^-4 and 8 x 10^-4? How did you know to go with -5 instead of -4 as the exponent? 

  5. Same question Jennifer, Natalie, and meowismeify asked below. You do not explain how you arrived at the exponent 5 instead of 4. I'm not sure how else to ask the question. What part of my questions don't you understand? The desired value lies between two exponents (4 and 5), and you picked 5. However, you do not provide a rationale for why you picked 5 instead of 4 or why you did not pick 4. As mentioned. the same question has been asked by 3 other people, but you haven't yet answered the question.

  6. Hi Leah, I really like your video very much and they help a lot! But I just have one question toward the end of the video: isn't 6.46 x 10^-5 actually closer to 5 x 10^-5 instead of closer to 8 x 10^-5 (I think you said it's closer to 8 x 10^-5)? In this case if there are 2 answer choices between 5 x 10^-5 and 8 x 10^-5 how do I know which one it is?

  7. Hello, love your videos, but I am very confused. I get up to the point where the Ka is in the range between 1×10^-4 and 1×10^-5 and am confused with the final answer you get. Isn't 6.46×10^-5 larger than 1×10^-5 therefore being out of the range??

  8. Basically what people are asking (I think), and which I'm also a little confused by, is at 4:27 where the answer is somewhere between 5×10^5 – 8×10^-5. I would have reasoned the range to instead be 5×10^4 – 8×10^4. I know this isn't right, but I don't see a logic behind why you chose the former (and correct) answer.

  9. Hey +Leah4sciMCAT,

    Great video! one question though. At 5:02 you fine tune your answer saying it should be closer to 8×10^-5, because .19 is closer to # .1 than to # .3, however the answer is actually closer to 5 x 10^-x. What there a mistake?

  10. Hello, I know you have tried to explain this before but I do not understand it. Why do you choose an answer between 5X10^-5 – 8X10^-5 instead of 5X10^-4 – 8×10^-4 ?

  11. At 4:15 mark I believe you misspoke when you said "point 1 gives me an 8 and point 5 gives me a 5". I think you meant to say "point 1 gives me an 8 and point 3 gives me a 5". Due to the fact that you rounded 4.19 to 4.2 and (.2) lies half way between (.1) and (.3), which would then give the (8 x 10^-x) and (5 x 10^-x) values you use later in the problem.

  12. i understood the first part now am confused how to calculate negative log of 5,7 or any decimal am not able to fig out
    please explain

  13. Hey Leah, I think you should delete this video and remake an updated version to cover some important points that you might have left out

  14. Hi Leah, why did you end up going for 10^-5 for the pH= 4.19 problem? We had two options of 10^-4 and 10^-5 when you explained that we need to round down and up to 4 and 5, respectively. but it wasn't explained why -5 was a better answer choice. Why did we go for 5?

  15. (1) 4.19+ 0.15=4.33 (Always add constant of 0.15 to your pKa value)
    (2) 5.00-4.33=0.67 (increase the given pKa to the whole next number and subtract from that whole number answer you get from step 1)
    (3) 0.67 x10 = 6.7 (always multiply whatever number you get by 10)
    (4) 10^-5 ( one thing to note your 10^- number will be alway one digit higher than given pKa value)
    (5) 6.7×10^-5
    Hope it helps everybody.
    Calc: 6.46×10^-5 .

  16. I think what people are confused is to why the -4 wasnt selected and why -5 was. Yes its because the pH would of have to been 3.19 but how can we calculate to know that? And eliminate -4.

  17. Leah 3:58 please explain why you chose 8 x 10^-5 instead of 8x 10^-4. I watched all the other videos in this series and took notes based on your videos, but I don't understand why you chose -5 instead of -4.

  18. I too was confused about the 10^5 thing but then I re-watched your last video again and it mostly makes sense.

    Rounding Ka up to -4 and down to -5:

    Ka of 5 gives a pKA of 1 x 10^-5
    Ka of 4 gives a pKA of 1 x 10^-4 ——> but really we should look at it as 10 x 10^-5

    So that's why we use 10^5, which is confusing because the Ka starts with a 4 (4.19)

    Now since 4.19 is effectively 4.2, we don't have the pKa value for the Coeffient of .2 (again, the .2 is from 4.2), but we have the values for

    .1 = 8 x 10^-x
    .3 = 5 x 10^-x

    So the final pKa answer would be between 5 x 10^-5 and 8 x 10^-5

    which is a much more narrow range than 1 x 10^-5 and 10 x 10^-5

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