How to find the number that the log was derived from? That is, how do you find the antilog? We will use the bicarbonate system as an example.

If you know the pH of a solution such as urine, blood, intracellular fluid, or intercellular fluid, and you know the pKa for a weak acid in that fluid, you can determine the ratio of the [conjugate base] /[weak acid].

_{3}

^{-}]/[H

_{2}CO

_{3}]

_{3}

^{-}]/[H

_{2}CO

_{3}]}.

Solve for [HCO_{3}^{-}]/[H_{2}CO_{3}] by inserting x into your calculator and depressing the 10^{X} button.

_{3}

^{-}]/[H

_{2}CO

_{3}]}, what is [HCO

_{3}

^{-}]/[H

_{2}CO

_{3}] equal to?

There are two ways to work this problem. The first is by remembering the 0.3 is the Log 2 so [HCO_{3}^{-}]/[H_{2}CO_{3}] = 2

The second way is to have a calculator with a 10^{X} button on it. You simply enter 0.3 and depress the 10^{X} button. The answer 2 (or 1.99999) should appear.

_{3}

^{-}]/[H

_{2}CO

_{3}], then what is [HCO

_{3}

^{-}]/[H

_{2}CO

_{3}].

There are two ways to work this problem. The first is to remember that the log of 10 is 1. If that is the case, then [HCO_{3}^{-}]/[H_{2}CO_{3}] must equal 10.

The second way is to enter 1 into your calculator and depress the 10^{X} button. 10 should appear as the answer.

_{3}

^{-}]/[H

_{2}CO

_{3}]}, then what would [HCO

_{3}

^{-}]/[H

_{2}CO

_{3}] be equal to?

If you do not like working with negative numbers, remember that –1= Log {[HCO_{3}^{-}]/[H_{2}CO_{3}]} is the same as 1 = Log {[H_{2}CO_{3}]/ [HCO_{3}^{-}]}.

Then there are two ways to work the problem. First, remember that the log of 10 is 1 so the ratio of [H_{2}CO_{3}]/ [HCO_{3}^{-}] must be 10. By inverting, the ratio of [HCO_{3}^{-}]/[H_{2}CO_{3}] = 0.1.

The second way is to enter -1 into your calculator and depress the 10^{X} button. 0.1 should appear as the answer.

1.6 = Log {[HCO_{3}^{-}]/[H_{2}CO_{3}]}, what is [HCO_{3}^{-}]/[H_{2}CO_{3}] equal to?

There are three ways to work this problem. The first is by changing the 1.6 into 1 + 0.3 + 0.3. Next, remember that the Log of xyz is equal to the Log x + Log y +Log z. So 1 + 0.3 + 0.3 is derived from the Log 10 X 2 X 2 = Log 40. So [HCO_{3}^{-}]/[H_{2}CO_{3}] = 10 X 2X 2 =40

The second way is to have calculator with a 10^{X} button on it. You simply enter 1.6 and depress the 10^{X} button. The answer 40 appears.

A third way is to remember that when the log x changes by 0.3, then x changes by a factor of 2.

If you had memorized that for normal blood, the values for the Hendersen-Hasselbach equation, pH = pKa + Log{[HCO_{3}^{-}]/[H_{2}CO_{3}]}, are 7.4 = 6.1 + Log [24mm]/[1.2 mm]. If the pH is increased by 7.3 to 7.6, that is , by 0.3, then the ratio [24mm]/[1.2 mm] had to increase by a factor of 2. that is, it had to increase from 20/1 to 40/1.