MCAT General Chemistry Review  Steven A. Leduc 2015
MCAT Math for General Chemistry
Logarithms
15.1 THE DEFINITION OF A LOGARITHM
A logarithm (or just log, for short) is an exponent.
For example, in the equation 2^{3} = 8, 3 is the exponent, so 3 is the logarithm. More precisely, since 3 is the exponent that gives 8 when the base is 2,
we say that the base2 log of 8 is 3, symbolized by the equation log_{2} 8 = 3.
Here’s another example: Since 10^{2} = 100, the base10 log of 100 is 2; that is, log_{10} 100 = 2. The logarithm of a number to a given base is the exponent the base needs to be raised to give the number. What’s the log, base 3, of 81? It’s the exponent we’d have to raise 3 to in order to give 81. Since 3^{4} = 81, the base3 log of 81 is 4, which we write as log_{3} 81 = 4.
The exponent equation 2^{3} = 8 is equivalent to the log equation log_{2} 8 = 3; the exponent equation 10^{2} = 100 is equivalent to the log equation log_{10} 100 = 2; and the exponent equation 3^{4} = 81 is equivalent to the log equation log_{3} 81 = 4. For every exponent equation, b^{x} = y, there’s a corresponding log equation: log_{b} y = x, and vice versa. To help make the conversion, use the following mnemonic, called the two arrows method:
You should read the log equations with the two arrows like this:
Always remember: The log is the exponent.
15.2 LAWS OF LOGARITHMS
There are only a few rules for dealing with logs that you’ll need to know, and they follow directly from the rules for exponents (given earlier, in Chapter 1). After all, logs are exponents.
In stating these rules, we will assume that in an equation like log_{b} y = x, the base b is a positive number that’s different from 1, and that y is positive. (Why these restrictions? Well, if b is negative, then not every number has a log. For example, log_{—3} 9 is 2, but what is log_{—3} 27? If b were 0, then only 0 would have a log; and if b were 1, then every number x could equal log_{1} y if y =1, and no number x could equal log_{1} y if y ≠ 1. And why must y be positive? Because if b is a positive number, then b^{x} [which is y] is always positive, no matter what real value we use for x. Therefore, only positive numbers have logs.)
Laws of Logarithms 

Law 1 
The log of a product is the sum of the logs: 
Law 2 
The log of a quotient is the difference of the logs: 
Law 3 
The log of (a number to a power) is that power times the log of the number: 
We could also add to this list that the log of 1 is 0, but this fact just follows from the definition of a log: Since b^{0} = 1 for any allowed base b, we’ll always have log_{b} 1 = 0.
For the MCAT, the two most important bases are b = 10 and b = e. Base10 logs are called common logs, and the “10” is often not written at all:
log y means log_{10} y
The base10 log is useful because we use a decimal number system, which is based (pun intended) on the number 10. For example, the number 273.15 means (2 × 10^{2}) + (7 × 10^{1}) + (3 × 10^{0}) + (1 × 10^{−1}) + (5 × 10^{−2}). In physics, the formula for the decibel level of a sound uses the base10 log. In chemistry, the base10 log has many uses, such as finding values of the pH, pOH, pK_{a}, and pK_{b}.
Basee logs are known as natural logs. Here, e is a particular constant, approximately equal to 2.7. This may seem like a strange number to choose as a base, but it makes calculus run smoothly—which is why it’s called the natural logarithm—because (and you don’t need to know this for the MCAT) the only numerical value of b for which the function f(x) = b^{x} is its own derivative is b = e = 2.71828.… Basee logs are often used in the mathematical description of physical processes in which the rate of change of some quantity is proportional to the quantity itself; radioactive decay is a typical example. The notation “ln” (the abbreviation, in reverse, for natural logarithm) is often used to mean log_{e}:
ln y means log_{e} y
The relationship between the base10 log and the basee log of a given number can be expressed as ln y ≈ 2.3 log y. For example, if y = 1000 = 10^{3}, then ln 1000 ≈ 2.3 log 1000 = 2.3 × 3 = 6.9. You may also find it useful to know the following approximate values:
log 2 ≈ 0.3 
ln 2 ≈ 0.7 
log 3 ≈ 0.5 
ln 3 ≈ 1.1 
log 5 ≈ 0.7 
ln 5 ≈ 1.6 
Example 151:
a) What is log_{3} 9?
b) Find log_{5} (1/25).
c) Find log_{4} 8.
d) What is the value of log_{16} 4?
e) Given that log 5 ≈ 0.7, what’s log 500?
f) Given that log 2 ≈ 0.3, find log (2 × 10^{−6}).
g) Given that log 2 ≈ 0.3 and log 3 × 0.5, find log (6 × 10^{23}).
Solution:
a) log_{3} 9 = x is the same as 3^{x} = 9, from which we see that x = 2. So, log_{3} 9 = 2.
b) log_{5} (1/25) = x is the same as 5^{x} = 1/25 = 1/5^{2} = 5^{−2}, so x = —2. Therefore, log_{5} (1/25) = —2.
c) log_{4} 8 = x is the same as 4^{x} = 8. Since 4^{x} = (2^{2})^{x} = 2^{2x} and 8 = 2^{3}, the equation 4^{x} = 8 is the same as 2^{2x} = 2^{3}, so 2x = 3, which gives x = 3/2. Therefore, log_{4} 8 = 3/2.
d) log_{16} 4 = x is the same as 16^{x} = 4. To find x, you might notice that the square root of 16 is 4, so 16^{1/2} = 4, which means log_{16} 4 = 1/2. Alternatively, we can write 16^{x} as (4^{2})^{x} = 4^{2x} and 4 as 4^{1}. Therefore, the equation 16^{x} = 4 is the same as 4^{2x} = 4^{1}, so 2x = 1, which gives x = 1/2.
e) log 500 = log (5 × 100) = log 5 + log 100, where we used Law 1 in the last step. Since log 100 = log 10^{2} = 2, we find that log 500 ≈ 0.7 + 2 = 2.7.
f) log (2 × 10^{−6}) = log 2 + log 10^{−6}, by Law 1. Since log 10^{−6} = —6, we find that log (2 × 10^{−6}) ≈ 0.3 + (—6) = —5.7.
g) log (6 × 10^{23}) = log 2 + log 3 + log 10^{23}, by Law 1. Since log 10^{23} = 23, we find that log (6 × 10^{23}) ≈ 0.3 + 0.5 + 23 = 23.8.
Example 152: In each case, find y.
a) log_{2} y = 5
b) log_{2} y = —3
c) log y = 4
d) log y = 7.5
e) log y = —2.5
f) ln y = 3
Solution:
a) log_{2} y = 5 is the same as 2^{5} = y, so y = 32.
b) log_{2} y = —3 is the same as 2^{−3} = y, which gives y = 1/2^{3} = 1/8.
c) log y = 4 is the same as 10^{4} = y, so y = 10,000.
d) log y = 7.5 is the same as 10^{7.5} = y. We’ll rewrite 7.5 as 7 + 0.5, so y =10^{7+(0.5)} = 10^{7} × 10^{0.5}. Because 10^{0.5} = 10^{1/2} = , which is approximately 3, we find that y ≈ 10^{7} × 3 = 3 × 10^{7}.
e) log y = —2.5 is the same as 10^{−2.5} = y. We’ll rewrite —2.5 as —3 + 0.5, so y = 10^{−3+(0.5)} = 10^{−3} × 10^{0.5}. Because 10^{0.5} = 10^{1/2} = , which is approximately 3, we have that y ≈ 10^{−3} × 3 = 0.003.
f) ln y = 3 means log_{e} y = 3; this is the same as y = e^{3} (which is about 20).
Example 153: The definition of the pH of an aqueous solution is
pH = —log [H_{3}O^{+}] (or, simply, —log [H^{+}])
where [H_{3}O^{+}] is the hydronium ion concentration (in M).
Part I: Find the pH of each of the following solutions:
a) coffee, with [H_{3}O^{+}] = 8 × 10^{−6} M
b) seawater, with [H_{3}O^{+}] = 3 × 10^{−9} M
c) vinegar, with [H_{3}O^{+}] = 1.3 × 10^{−3} M
Part II: Find [H_{3}O^{+}] for each of the following pH values:
d) pH = 7
e) pH = 11.5
f) pH = 4.7
Solution:
a) pH = —log (8 × 10^{−6}) = —[log 8 + log (10^{−6})] = —log 8 + 6. We can now make a quick approximation by simply noticing that log 8 is a little less than log 10; that is, log 8 is a little less than 1. Let’s say it’s 0.9. Then pH ≈ —0.9 + 6 = 5.1.
b) pH = —log (3 × 10^{−9}) = —[log 3 + log (10^{−9})] = —log 3 + 9. We now make a quick approximation by simply noticing that log 3 is about 0.5 (after all, 9^{0.5} is 3, so 10^{0.5} is close to 3). This gives pH ≈ —0.5 + 9 = 8.5.
c) pH = —log (1.3 × 10^{−3}) = —[log 1.3 + log (10^{−3})] = —log 1.3 + 3. We can now make a quick approximation by simply noticing that log 1.3 is just a little more than log 1; that is, log 1.3 is a little more than 0. Let’s say it’s 0.1. This gives pH ≈ —0.1 + 3 = 2.9.
Note 1:
We can generalize these three calculations as follows: If [H_{3}O^{+}] = m × 10^{−n} M, where 1 ≤ m < 10 and n is an integer, then the pH is between (n — 1) and n; it’s closer to (n — 1) if m > 3 and it’s closer to n if m < 3. (We use 3 as the cutoff since log 3 ≈ 0.5.)
d) If pH = 7, then —log [H_{3}O^{+}] = 7, so log [H_{3}O^{+}] = —7, which means [H_{3}O^{+}] = 10^{−7} M.
e) If pH = 11.5, then —log [H_{3}O^{+}] = 11.5, so log [H_{3}O^{+}] = —11.5, which means [H_{3}O^{+}] = 10^{−11.5} = 10^{(0.5)—12} = 10^{0.5} × 10^{−12} ≈ 3 × 10^{−12} M.
f) If pH = 4.7, then —log [H_{3}O^{+}] = 4.7, so log [H_{3}O^{+}] = —4.7, which means [H_{3}O^{+}] = 10^{−4.7} = 10^{(0.3)—5} = 10^{0.3} × 10^{−5} ≈ 2 × 10^{−5} M. [10^{−0.3} ≈ 2 follows from the fact that log 2 ≈ 0.3.]
Note 2:
We can generalize these last two calculations as follows: If pH = n.m, where n is an integer and m is a digit from 1 to 9, then [H_{3}O^{+}] = y × 10^{−(n+1)} M, where y is closer to 1 if m > 3 and closer to 10 if m < 3. (We take y = 5 if m = 3.)
Example 154: The definition of the pK_{a} of a weak acid is
pK_{a} = —log K_{a}
where K_{a} is the acid’s ionization constant.
Part I: Approximate the pK_{a} of each of the following acids:
a) HBrO, with K_{a} = 2 × 10^{−9}
b) HNO_{2}, with K_{a} = 7 × 10^{−4}
c) HCN, with K_{a} = 6 × 10^{−10}
Part II: Approximate K_{a} for each of the following pK_{a} values:
d) pK_{a} = 12.5
e) pK_{a} = 2.7
f) pK_{a} = 9.2
Solution:
a) pK_{a} = —log (2 × 10^{−9}) = —[log 2 + log (10^{−9})] = —log 2 + 9. We can now make a quick approximation by remembering that log 2 is about 0.3. Then pK_{a} = —0.3 + 9 = 8.7. Because the formula to find pK_{a} from K_{a} is exactly the same as the formula for finding pH from [H^{+}], we could also make use of Note 1 in the solution to Example 153. If K_{a} = m´ × 10^{−}^{n} M, where 1 ≤ m < 10 and n is an integer, then the pK_{a} is between (n — 1) and n; it’s closer to (n — 1) if m > 3 and it’s closer to n if m < 3. In this case, m = 2 and n = 9, so the pK_{a} is between (n — 1) = 8 and n = 9. And, since 2 < 3, the pK_{a} will be closer to 9 (which is just what we found, since we got the value 8.7). Given a list of possible choices for the pK_{a} of this acid, just recognizing that it’s a little less than 9 will be sufficient.
b) With K_{a} = 7 × 10^{−4}, we have m = 7 and n = 4. Therefore, the pK_{a} will be between (n — 1) = 3 and n = 4. Since m = 7 is greater than 3, the value of pK_{a} will be closer to 3 (around, say, 3.2).
c) With K_{a} = 6 × 10^{−10}, we have m = 6 and n = 10. Therefore, the pK_{a} will be between (n — 1) = 9 and n = 10. Since m = 6 is greater than 3, the value of pK_{a} will be closer to 9 (around, say, 9.2).
d) If pK_{a} = 12.5, then —log K_{a} = 12.5, so log K_{a} = —12.5, which means K_{a} = 10^{−12.5} = 10^{(0.5)—13} = 10^{0.5} × 10^{−13} ≈ 3 × 10^{−13}. We could also make use of Note 2 in the solution to Example 153. If pK_{a} = n.m, where n is an integer and m is a digit from 1 to 9, then K_{a} = y × 10^{−(n+1)} M, where y is closer to 1 if m > 3 and y is closer to 10 if m < 3. In this case, with pK_{a} = 12.5, we have n = 12 and m = 5, so the K_{a} value is y × 10^{−(12+1)} = y × 10^{−13}, with y closer to 1 (than to 10) since m = 5 is greater than 3 (this agrees with what we found, since we calculated that K_{a} ≈ 3 × 10^{−13}).
e) With pK_{a} = 2.7, we have n = 2 and m = 7. Therefore, the K_{a} value is y × 10^{−(2+1)} = y × 10^{−3}, with y close to 1 since m = 7 is greater than 3. We can check this as follows: If pK_{a} = 2.7, then —log K_{a} = 2.7, so log K_{a} = —2.7, which means K_{a} = 10^{−2.7} = 10^{(0.3)—3} = 10^{0.3} × 10^{−3} ≈ 2 × 10^{−3}.
f) With pK_{a} = 9.2, we have n = 9 and m = 2. Therefore, the K_{a} value is y × 10^{−(9+1)} = y × 10^{−10}, with y closer to 10 (than to 1) since m = 2 is less than 3. We can say that K_{a} ≈ 6 × 10^{−10}.
Example 155:
a) If y increases by a factor of 100, what happens to log y?
b) If y decreases by a factor of 1000, what happens to log y?
c) If y increases by a factor of 30,000, what happens to log y?
d) If y is reduced by 99%, what happens to log y?
Solution:
a) If y changes to y´ = 100y, then the log increases by 2, since
log y´ = log(100y) = log 100 + log y = log 10^{2} + log y = 2 + log y
b) If y changes to y´ = y/1000, then the log decreases by 3, since
log y´ = log() = log y − log 1000 = log y − log 10^{3} = log y − 3
c) If y changes to y´ = 30,000y, then the log increases by about 4.5, since
log y´ = log(30000 y) = log 3 + log 10000 + log y ≈ 0.5 + 4 + log y = 4.5 + log y
d) If y is reduced by 99%, that means we’re subtracting 0.99y from y, which leaves 0.01y = y/100. Therefore, y has decreased by a factor of 100. And if y changes to y´ = y/100, then the log decreases by 2, since
log y´ = log () = log y − log 100 = log y − log 10^{2} = log y − 2
Example 156: A radioactive substance has a halflife of 70 hours. For each of the fractions below, figure out how many hours will elapse until the amount of substance remaining is equal to the given fraction of the original amount.
a) 1/4
b) 1/8
c) 1/3
Solution:
a) After one halflife has elapsed, the amount remaining is 1/2 the original (by definition). After another halflife elapses, the amount remaining is now 1/2 of 1/2 the original amount, which is 1/4 the original amount. Therefore, a decrease to 1/4 the original amount requires 2 halflives, which in this case is 2(70 hr) = 140 hr.
b) The fraction 1/8 is equal to 1/2 of 1/2 of 1/2; that is, 1/8 = (1/2)^{3}. In terms of halflives, a decrease to 1/8 the original amount requires 3 halflives, which in this case is equal to 3(70 hr) = 210 hr. In general, a decrease to (1/2)^{n} the original amount requires n halflives.
c) The fraction 1/3 is not a wholenumber power of 1/2, so we can’t directly apply the fact given in the italicized sentence in the solution to part (b). However, 1/3 is between 1/2 and 1/4, so the time to get to 1/3 the original amount is between 1 and 2 halflives. Since one halflife is 70 hr, the amount of time is between 70 and 140 hours; the middle of this range (since 1/3 is roughly in the middle between 1/2 and 1/4) is about 110 hours. The most general formula for calculating the elapsed time involves a logarithm: If x < 1 is the fraction of a radioactive substance remaining after a time t has elapsed, then
where t_{1/2} is the halflife. (If you want to use this formula, remember that log 2 ≈ 0.3.)