Dear forum,
I have been a long time reader of this forum and got a lot out of reading the discussions here thus far, so I thought I would give back and explain the method I use for logarithm calculation when solving for pH's and pOH's. The popular method described by most study books (herein referred to as the "estimation method"), when given -log(n*10^-m), is to estimate the answer as being m if n=1, being greater than m.5 if n is greater than ~3, and being less than m.5 if n is less than ~3. This gives you some estimated result that you compare to the multiple choices, from which you then choose the best answer as being the one closest to your estimated result.
I want to explain the way I approach the calculation of -log(n*10^-m), which I will call the "exact method" herein, because it affords the following benefits over the classic estimation method:
1. The exact method gives you an exact answer, with an error margin of at most +/- 0.05 when calculating pH/pOH (compared to a 10x higher error margin of 0.5 with the estimation method). I.e. the exact method will give you a result of, say, pH=5.3 whereas the estimation method gives you the result of "pH is greater than 5, and less than 6").
2. Because you get an exact answer, you can directly compare this exact answer to the multiple choice options and thus either (a) feel confident in your answer if it is exactly listed in the multiple choices, or (b) Spot a potential error in your calculations if your answer is more than ~0.1-0.2 away from the listed answers for the pH or pOH of a solution.
3. Even though you get a much, much more precise answer than with the estimation method, the exact method is equally fast or takes at most 2-5 seconds longer. I know you might be skeptical of this claim, but hear me out, and come to your own conclusion.
The only downside of the exact method is that you must memorize short list of values, but luckily the majority are whole numbers, and as you will see, the numbers themselves follow a very convenient sequence and will take you at most ~10 minutes to memorize. Practice a few pH problems with this method, and the table of values will be permanently ingrained into memory.
Now, onto the method itself.
We first have the definition of a logarithm:
The output of log(n) gives you the power that 10 would have to be raised to in order to equal n. That is, log(1000)=3, since 10^3 = 1000. Similarly, log(3.16) = 0.5, since 10^0.5 = 3.16. And so on. (If you don't know this notation, the " ^ " symbol means "to the power of." )
We also know the following important rule when computing logarithms:
log(A*B) = log(A) + log(b)
From the second rule we see, if we're given a number like n*10^-m, that:
-log(n*10^-m) = -[log👎 + log(10^-m)] = -log(n)+m = m-log(n)
As you will see, the only difference between the estimation method and this exact method is that in the estimation method, we say the answer is m minus some decimal value, whereas with the exact method, we can directly calculate the value of m-log(n) in less than a second and thus get an exact answer for the pH instead of resorting to just "m minus some decimal value."
Firstly, from the formula -log(n*10^-m) = m-log(n) we know that n will always be greater than 1 but less than 10 (since if n=1 then pH=m, and if n>10 or n<1 then we can just move the decimal point and change the value of -m accordingly--i.e. 50x10^-4 = 5x10^-3 and 0.5x10^-2 = 5x10^-3).
Since we know that 1<n<10, if we find a table of values z such that log(n)=z for n=1, 2, 3, ...9, then we can immediately calculate the value of log(n) without having to use a calculator.
So far, this may not sound too easy and/or seem like an overly involved process, but it is about to get a whole lot easier and make a whole lot more sense.
The following table is ALL you need to remember for this method. If you write it down 4-5 times, you'll have it memorized:
10^0.1 = 1.25
10^0.2 = 1.6
10^0.3 = 2
10^0.4 = 2.5
10^0.5 = 3.16
10^0.6 = 4
10^0.7 = 5
10^0.8 = 6.3
10^0.85 = 7
10^0.9 = 8
10^0.95 = 9
Another way to write this table would be as follows, for the same result:
0.1 = log(1.25)
0.2 = log(1.6)
0.3 = log(2)
0.4 = log(2.5)
0.5 = log(3.16)
0.6 = log(4)
0.7 = log(5)
0.8 = log(6.3)
0.85 = log(7)
0.9 = log(8)
0.95 = log(9)
Again, after I rewrote this from memory 5-6 times, I had it memorized. What happens if you forget an entry? Well, it's very easy to fill in missing values.
For example, what if we forgot that log(5) = 0.7?
Well, since log(A*B) = log(A) + log(B) we'd know that log(5) = log(2*2.5) = log(2) + log(2.5) = 3 + 4 = 7, so log (5) = 7, just like it says in the table.
Now let's look at some example problems that demonstrate how this exact method is useful:
1. Find the pH of a 0.007M solution of HCl:
The [H+] concentration would be 0.007M (since HCl dissociates completely), so:
pH = -log(0.007)
pH = -log(7x10^-3)
pH = 3 - log(7) = 3 - 0.85 (after recalling from the memorized table that log(7) = 0.85)
pH = 2.15
2. Find the pH of a 0.2mM solution of NaOH:
The [OH-] concentration would be 0.2mM = 0.0002M (since NaOH dissociates completely), so:
pOH = -log(0.0002)
pOH = -log(2x10^-4)
pOH = 4 - log(2) = 4 - 0.3 (after recalling from the memorized table that log(2) = 0.3)
pOH = 3.7
Then, assuming 25 deg. C temperature,
pH = 14 - pOH = 14 - 3.7
pH = 10.3
Consider that if you did this using the classic "estimation method," you'd have to think, "Well, the negative log of 2x10^-4 is more than 3 but less than 4, and since 2 is less than 3.16, then the negative log must be greater than 3.5 but less than 4, and so to find pH we have to take 14 and subtract a number greater than 3.5 but less than 4, which means that pH is between 10 and 10.5." Since after trying the exact method two or three times you will instantly know that -log(2x10^-4) = 4-log(2) = 4-0.3 = 3.7 and thus 14 - 3.7 = 10.3 (and doing this in your head will quickly become faster than logically working with vague ranges as per the estimation method), the exact method will get you an exact answer and save you time over the estimation method that produces a more vague solution.
(Keep in mind that while you're taking the real MCAT you will undoubtedly experience somewhat more stress/anxiety than you would when you're taking a practice test. So even if you're somewhat comfortable with juggling multiple ranges/numbers in your head as you calculate the pH of a solution this way, when it's time to actually do this estimation in your head during the real MCAT, you're more likely to slip up with an error or doubt yourself by thinking, "wait, what range did I say the pOH had again?" than you are to slip up when doing a subtraction with cold, hard numbers. It's easier to perform "14 - 3.7" in your head than it is to perform "[/i]14 minus some number bigger than 3.5 but less than 4[/i]" in your head.)
3. Find the pH of a solution that has an [H+] concentration of 7.5x10^-3.
pH = -log(7.5x10^-3)
pH = 3 - log(7.5)
pH = 3 - 0.87 (since 7.5 is between 7 and 8 on the chart, so log(7.5) should be ~halfway between log(7) and log(8) )
pH = 2.13
Quick, quick answers.
By getting a much more exact answer, you also save the time you would otherwise spend comparing the vague range resulting from the estimation method to the various multiple choice answers given. You simply choose the multiple choice answer that is the same as your result from the exact method, rather than thinking, "well, there are two answers here that might work, let's now think further into which one is the better answer..."
The table you need to memorize may be simply committed to memory by just memorizing the following number pairs:
(0.1 , 1.25)
(0.2 , 1.6)
(0.3 , 2)
(0.4 , 2.5)
(0.5 , 3.16)
(0.6 , 4)
(0.7 , 5)
(0.8 , 6.3)
(0.85 , 7)
(0.9 , 8)
(0.95 , 9)
From which the log of the number on the right equals the number on the left.
I hope this post has been useful! The classic estimation method is definitely also viable as a way to get your answer, but I just found that this "exact method" is quicker and also gives far more precise answers that are (for me) much easier to work with, whether I am doing the math in my head or on paper.
Cheers, and let me know what you all think!
I have been a long time reader of this forum and got a lot out of reading the discussions here thus far, so I thought I would give back and explain the method I use for logarithm calculation when solving for pH's and pOH's. The popular method described by most study books (herein referred to as the "estimation method"), when given -log(n*10^-m), is to estimate the answer as being m if n=1, being greater than m.5 if n is greater than ~3, and being less than m.5 if n is less than ~3. This gives you some estimated result that you compare to the multiple choices, from which you then choose the best answer as being the one closest to your estimated result.
I want to explain the way I approach the calculation of -log(n*10^-m), which I will call the "exact method" herein, because it affords the following benefits over the classic estimation method:
1. The exact method gives you an exact answer, with an error margin of at most +/- 0.05 when calculating pH/pOH (compared to a 10x higher error margin of 0.5 with the estimation method). I.e. the exact method will give you a result of, say, pH=5.3 whereas the estimation method gives you the result of "pH is greater than 5, and less than 6").
2. Because you get an exact answer, you can directly compare this exact answer to the multiple choice options and thus either (a) feel confident in your answer if it is exactly listed in the multiple choices, or (b) Spot a potential error in your calculations if your answer is more than ~0.1-0.2 away from the listed answers for the pH or pOH of a solution.
3. Even though you get a much, much more precise answer than with the estimation method, the exact method is equally fast or takes at most 2-5 seconds longer. I know you might be skeptical of this claim, but hear me out, and come to your own conclusion.
The only downside of the exact method is that you must memorize short list of values, but luckily the majority are whole numbers, and as you will see, the numbers themselves follow a very convenient sequence and will take you at most ~10 minutes to memorize. Practice a few pH problems with this method, and the table of values will be permanently ingrained into memory.
Now, onto the method itself.
We first have the definition of a logarithm:
The output of log(n) gives you the power that 10 would have to be raised to in order to equal n. That is, log(1000)=3, since 10^3 = 1000. Similarly, log(3.16) = 0.5, since 10^0.5 = 3.16. And so on. (If you don't know this notation, the " ^ " symbol means "to the power of." )
We also know the following important rule when computing logarithms:
log(A*B) = log(A) + log(b)
From the second rule we see, if we're given a number like n*10^-m, that:
-log(n*10^-m) = -[log👎 + log(10^-m)] = -log(n)+m = m-log(n)
As you will see, the only difference between the estimation method and this exact method is that in the estimation method, we say the answer is m minus some decimal value, whereas with the exact method, we can directly calculate the value of m-log(n) in less than a second and thus get an exact answer for the pH instead of resorting to just "m minus some decimal value."
Firstly, from the formula -log(n*10^-m) = m-log(n) we know that n will always be greater than 1 but less than 10 (since if n=1 then pH=m, and if n>10 or n<1 then we can just move the decimal point and change the value of -m accordingly--i.e. 50x10^-4 = 5x10^-3 and 0.5x10^-2 = 5x10^-3).
Since we know that 1<n<10, if we find a table of values z such that log(n)=z for n=1, 2, 3, ...9, then we can immediately calculate the value of log(n) without having to use a calculator.
So far, this may not sound too easy and/or seem like an overly involved process, but it is about to get a whole lot easier and make a whole lot more sense.
The following table is ALL you need to remember for this method. If you write it down 4-5 times, you'll have it memorized:
10^0.1 = 1.25
10^0.2 = 1.6
10^0.3 = 2
10^0.4 = 2.5
10^0.5 = 3.16
10^0.6 = 4
10^0.7 = 5
10^0.8 = 6.3
10^0.85 = 7
10^0.9 = 8
10^0.95 = 9
Another way to write this table would be as follows, for the same result:
0.1 = log(1.25)
0.2 = log(1.6)
0.3 = log(2)
0.4 = log(2.5)
0.5 = log(3.16)
0.6 = log(4)
0.7 = log(5)
0.8 = log(6.3)
0.85 = log(7)
0.9 = log(8)
0.95 = log(9)
Again, after I rewrote this from memory 5-6 times, I had it memorized. What happens if you forget an entry? Well, it's very easy to fill in missing values.
For example, what if we forgot that log(5) = 0.7?
Well, since log(A*B) = log(A) + log(B) we'd know that log(5) = log(2*2.5) = log(2) + log(2.5) = 3 + 4 = 7, so log (5) = 7, just like it says in the table.
Now let's look at some example problems that demonstrate how this exact method is useful:
1. Find the pH of a 0.007M solution of HCl:
The [H+] concentration would be 0.007M (since HCl dissociates completely), so:
pH = -log(0.007)
pH = -log(7x10^-3)
pH = 3 - log(7) = 3 - 0.85 (after recalling from the memorized table that log(7) = 0.85)
pH = 2.15
2. Find the pH of a 0.2mM solution of NaOH:
The [OH-] concentration would be 0.2mM = 0.0002M (since NaOH dissociates completely), so:
pOH = -log(0.0002)
pOH = -log(2x10^-4)
pOH = 4 - log(2) = 4 - 0.3 (after recalling from the memorized table that log(2) = 0.3)
pOH = 3.7
Then, assuming 25 deg. C temperature,
pH = 14 - pOH = 14 - 3.7
pH = 10.3
Consider that if you did this using the classic "estimation method," you'd have to think, "Well, the negative log of 2x10^-4 is more than 3 but less than 4, and since 2 is less than 3.16, then the negative log must be greater than 3.5 but less than 4, and so to find pH we have to take 14 and subtract a number greater than 3.5 but less than 4, which means that pH is between 10 and 10.5." Since after trying the exact method two or three times you will instantly know that -log(2x10^-4) = 4-log(2) = 4-0.3 = 3.7 and thus 14 - 3.7 = 10.3 (and doing this in your head will quickly become faster than logically working with vague ranges as per the estimation method), the exact method will get you an exact answer and save you time over the estimation method that produces a more vague solution.
(Keep in mind that while you're taking the real MCAT you will undoubtedly experience somewhat more stress/anxiety than you would when you're taking a practice test. So even if you're somewhat comfortable with juggling multiple ranges/numbers in your head as you calculate the pH of a solution this way, when it's time to actually do this estimation in your head during the real MCAT, you're more likely to slip up with an error or doubt yourself by thinking, "wait, what range did I say the pOH had again?" than you are to slip up when doing a subtraction with cold, hard numbers. It's easier to perform "14 - 3.7" in your head than it is to perform "[/i]14 minus some number bigger than 3.5 but less than 4[/i]" in your head.)
3. Find the pH of a solution that has an [H+] concentration of 7.5x10^-3.
pH = -log(7.5x10^-3)
pH = 3 - log(7.5)
pH = 3 - 0.87 (since 7.5 is between 7 and 8 on the chart, so log(7.5) should be ~halfway between log(7) and log(8) )
pH = 2.13
Quick, quick answers.
By getting a much more exact answer, you also save the time you would otherwise spend comparing the vague range resulting from the estimation method to the various multiple choice answers given. You simply choose the multiple choice answer that is the same as your result from the exact method, rather than thinking, "well, there are two answers here that might work, let's now think further into which one is the better answer..."
The table you need to memorize may be simply committed to memory by just memorizing the following number pairs:
(0.1 , 1.25)
(0.2 , 1.6)
(0.3 , 2)
(0.4 , 2.5)
(0.5 , 3.16)
(0.6 , 4)
(0.7 , 5)
(0.8 , 6.3)
(0.85 , 7)
(0.9 , 8)
(0.95 , 9)
From which the log of the number on the right equals the number on the left.
I hope this post has been useful! The classic estimation method is definitely also viable as a way to get your answer, but I just found that this "exact method" is quicker and also gives far more precise answers that are (for me) much easier to work with, whether I am doing the math in my head or on paper.
Cheers, and let me know what you all think!
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