Here are four pH problems and a few hints that, for the most part, were shown on the board during the lecture.

Given any hydrogen ion concentration, determine the pH and visa versa.

First some useful conversions:

Log 2 = 0.30103

Log xy = Log x + Log y

So, Log 4 = Log [2 X 2] = Log 2 + Log 2 = 0.6

Likewise, Log 8 = Log 2 + Log 2 + Log 2 = 0.9

And, another conversion

Log xy = Log x + Log y

So, Log 4 = Log [2 X 2] = Log 2 + Log 2 = 0.6

Likewise, Log 8 = Log 2 + Log 2 + Log 2 = 0.9

And, another conversion

Log x = -Log 1/x

Log x/y = -Log y/x

So Log 0.5 = Log [1/2] = -Log 2 = -0.3

**Example 1**: The [H

^{+}] of pure water is 1 X 10

^{-7}mol/L, calculate the pH.

pH = -Log [H

^{+}] = -Log (1 X 10^{-7}) = -Log 1 + -Log 10^{-7}= -0 + 7 = 7**Examples 2, 3, and 4**: The determination the pH of different blood samples from the [H

^{+}].

The [H

pH = - Log [H

= -0.6 + 8 = 7.4

The hydrogen ion is doubled to 80 nM, calculate the pH.

pH = - Log [H

= -0.9 + 8 = 7.1

The hydrogen ion is halved to 20 nM, calculate the pH.

pH = - Log [H

= -0.3 + 8 = 7.7

^{+}] of normal blood is about 40 nM (nanomolar), calculate the pH.pH = - Log [H

^{+}] = -Log [40 X 10^{-9}] = -Log[ 4 X 10^{-8}] = -Log 4 – Log 10^{-8}= -0.6 + 8 = 7.4

The hydrogen ion is doubled to 80 nM, calculate the pH.

pH = - Log [H

^{+}] = -Log [80 X 10^{-9}] = -Log[ 8 X 10^{-8}] = -Log 8 – Log 10^{-8}= -0.9 + 8 = 7.1

The hydrogen ion is halved to 20 nM, calculate the pH.

pH = - Log [H

^{+}] = -Log [20 X 10^{-9}] = -Log[ 2 X 10^{-8}] = -Log 2 – Log 10^{-8}= -0.3 + 8 = 7.7

Notice that whenever we change the pH by 0.3, we are changing the hydrogen ion concentration by a factor of 2.

When you are given the log and asked for the number that the log was derived from, work the problem backwards. Or, for the last problem, enter -7.7 into your calculator and pres the 10^{X} button.