Here's the scenario:

a) You're given a hydrogen ion concentration (say 1.00 x 10¯^{8}M) and you are asked for the pH.

b) You perform the calculation (-log 1.00 x 10¯^{8}).

c) You announce the answer: 8.000

d) To your horror, you are told this is the wrong answer.

What in the world happened?

The answer is that the negative log technique is actually an approximation, but you're not told that during the usual teaching of a chemistry class. On a personal note, the ChemTeam does not remember learning this particular trick during his chemistry classes in high school and college. (Of course, those classes were in the "ancient" days of the late 1960's and early 1970's, so 'nuff said!)

The approximation is that the contribution of hydrogen ion from water is ignored. After all, pure water already contains hydrogen ion at a concentration of 1.00 x 10¯^{7} M. When the hydrogen ion contribution from an acid is 1000 to 1,000,000 times more, the contribution from water can safely be ignored.

Not so when the [H^{+}] = 1.00 x 10¯^{8} M. In that case, a more complex technique is used which takes the hydrogen ion from water into account. When that is done, the correct answer to the above example is 6.98. However, that more complex technique will not be discussed here.

By the by, I discuss the above example trick question in a video. Here it is.

**Problem #1:** Here's a variant that's not framed in a tricky way:

Show that the pH of a solution remains 7.00 when 1.0 x 10¯^{11}moles of HCl is added to the solution.

**Solution:**

1) pH of 7.00 means this:

[H^{+}] = 1.0 x 10¯^{7}M

2) Add 1.0 x 10¯^{11} moles of H^{+} to get:

1.0001 x 10¯^{7}MpH = - log 1.0001 x 10¯

^{7}= 6.9999566

3) Round the answer off to two significant figures:

pH = 7.00

By the way, no pH meter constructed is sensitive enough to determine a value like 6.9999566. Heck, it's a tricky business to get 3 or 4 sig figs on a pH meter, much less 6 or 7.

**Problem #2:** Another variation:

Assume that you prepared a solution by adding 0.05 mL (1 drop) of 0.500 HCl to 5.00 x 10^{7}mL of "pure" water. Calculate the resultant pH. (Hint: the answer is not pH = 9.3; add up the hydronium ions from ALL sources)

**Solution:**

The key is to remember that "pure" water is naturally 1.00 x 10¯^{7} M in hydrogen ion. That's the thinking behind the "ALL sources" hint

1) Calculate total H^{+} in 5.00 x 10^{7} mL of water:

(1.00 x 10¯^{7}mol/L) (5.00 x 10^{4}L) = 0.005 mol

2) Calculate total H^{+} in HCl:

(0.500 mol/L) (0.00005 L) = 0.000025 mol

3) Add the results together:

0.005 mol + 0.000025 mol = 0.005025 mol

4) Calculate new molarity of hydrogen ion:

0.005025 mol / 5.00 x 10^{4}L = 1.005 x 10¯^{7}M

5) Calculate new pH:

- log 1.005 x 10¯^{7}= 6.9978

Rounded off to three sig figs, the pH is 6.998. Since we had 0.05 mL, let's round off to one sig fig to arrive at a pH of 7.0.

By the way, a pH of 9.3 is arrived at like this:

H^{+}from HCl = 0.000025 mol

total volume = 5.00 x 10^{4}L[H

^{+}] = 0.000025 mol / 5.00 x 10^{4}L = 5.00 x 10¯^{10}MpH = -log 5.00 x 10¯

^{10}= 9.301

**Problem #3:** Another variation someday? One can hope!