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 usually not told that during the teaching of the pH concept.

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.

**Example #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.

**Example #2:** Another variation:

Assume that you prepared a solution by adding 0.050 mL (the usual volume for 1 drop) of 0.500 M HCl to 5.00 x 10^{7}mL of "pure" water. Calculate the resultant pH. (Hint: the answer is not pH = 9.30 or pH = −0.30; 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 "pure" water:

(1.00 x 10¯^{7}mol/L) (5.00 x 10^{4}L) = 0.0050 mol

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

(0.500 mol/L) (0.000050 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 / 5.000000005 x 10^{4}L = 1.005 x 10¯^{7}MFor the final volume, I added 0.050 mL and 5.00 x 10

^{7}mL and then converted to liters.)

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 0.050 mL, let's round off to two sig figs to arrive at a pH of 7.00.

By the way, the wrong pH of 9.30 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}= 9.30

I will leave you to ponder how the wrong answer of −0.30 is arrived at.

**Example #3:** Calculate the pH of a 1.30 x 10¯^{9} M of HBr solution

**Solution:**

HBr is a strong acid. Each mole of HBr ionizes 100% to give 1 mole of H_{3}O^{+}.Therefore, [H

_{3}O^{+}] from HBr = 1.30 x 10¯^{9}MIf one calculates the pH from the information above, a value of 8.886 is obtained. This is a basic pH and cannot be obtained by putting an acid (the HBr) into solution.

In most pH problems, you ignore the autoionization of water that produces [H

_{3}O^{+}] = 1.00 x 10¯^{7}M. This is done because we are usually dealing with solutions of acids and bases of much higher concentrations and the autoionization of water is relatively insignificant at those higher concentrations. But, when dealing with very low concentrations of acids and bases (ones that are very close to 1.00 x 10¯^{7}M), the autoionization of water is no longer insignificant and must be included in the calculation.

2H _{2}O(ℓ)⇌ H _{3}O^{+}(aq)+ OH¯(aq) K _{w}= 1.00 x 10¯^{14}Initial (M): -- 1.30 x 10¯ ^{9}0 Change (M): -- +y +y Equil (M): -- (1.30 x 10¯ ^{9}+ y)y At equilibrium:

K_{w}= [H_{3}O^{+}] [OH¯]1.00 x 10¯

^{14}= (1.30 x 10¯^{9}+ y) (y)y

^{2}+ (1.30 x 10¯^{9}) (y) − (1.0 x 10¯^{14}) = 0$y=\frac{-\mathrm{1.30\; x\; 10\xaf9}\pm \sqrt{{\mathrm{(1.30\; x\; 10\xaf9)2}}^{\mathrm{}}+\mathrm{(4)}\mathrm{(1)}\mathrm{(1.0\; x\; 10\xaf14)}}}{2\mathrm{(1)}}$

y = 9.94 x 10¯

^{8}(after rejecting the other root of −1.01 x 10¯^{7}[H

_{3}O^{+}] = (1.30 x 10¯^{9}+ 9.94 x 10¯^{8}) M = 1.007 x 10¯^{7}MpH = −log [H

_{3}O^{+}] = −log (1.007 x 10¯^{7}) = 6.997