Repeated Figures From Chapter 4

Here are five problems that, for the most part, were shown on the board during the lecture. Concepts covered are the ion product of water, the pH, and the Hendersen-Hasselbach equation.

Define the ion product for water! Given any hydroxide ion concentration, determine the hydrogen ion concentration and visa versa. Note this problem was not shown in class.
The [H+] of a solution is 5 X 10-5 M. What is the [OH-]?
The brackets designate concentration.

Kw = 1 X 10-14 = [H+] [OH-] = 5 X 10-5 [OH-]

Since 1 X 10-14 = 10 X 10-15

[OH-] = (10 X 10-15) / (5 X 10-5) = 2 X 10-10

To check the problem:

(5 X 10-5)(2 X 10-10) = 10 X 10-15 = 1 X 10-14
Define pH. 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 x = -Log 1/x

Log x/y = -Log y/x

So Log 0.5 = Log [1/2] = -Log2 = -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+] 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, enter -7.7 into your calculator and pres the 10X button.
Be able to draw the equations that show how the bicarbonate buffer system works in blood. What is the respiratory compensation when the blood pH drops to 7.3? What is the respiratory compensation when the blood pH rises to 7.5?
CO2 + H2O = H2CO3 = H+ + HCO3-

pH = pKa + Log [HCO3-]/[H2CO3]

If metabolism produces excess acid (H+), the H+ combines with and lowers the concentration of bicarbonate ([HCO3-]).

If excess base is added, carbonic acid dissociates to provide protons to neutralize the base. As a result, the concentration of bicarbonate increases.

If the bicarbonate buffering system causes a change in the pH, respiratory compensation will occur. Breathing faster and deeper will remove carbonic acid and more shallow breathing will increase the concentration of carbonic acid. This is compensatory respiratory alkalosis and acidosis respectively. Note that in both cases, the pH moves back toward 7.4.

Note that PaCO2 is the partial pressure of CO2 in arterial blood. It is reported as mmHg and can be converted to mM by multiplying by 0.03. In this class, the units for the concentrations in the Henderson-Hasselbalch equation will always be in mM (mmol/L) (meq/ml).

Remember that the [H2CO3] is regulated by the brain and lungs.

You should memorize the following normal values for the bicarbonate buffer system in a normal person:

pH = pKa + Log [HCO3-]/[H2CO3]

7.4 = 6.1 + Log [24mM]/[1.2mM]

Notice that if you are given any pH and pKa, you can determine the ratio of [HCO3-]/[H2CO3]
Draw the dissociation of ammonium ion.

If the pKa =9.25, what form is found at pH = 7.4?
NH4+ = H+ + NH3

pH = pKa + Log [NH3]/[NH4+]

7.4 = 9.25 + Log [NH3]/[NH4+]

-1.85 = Log [NH3]/[NH4+]

1.85 = Log [NH4+]/[NH3]

71 = [NH4+]/[NH3]

Ammonium ion predominates.
Concerning Dennis Veere: If the pKa for acetylsalicylic acid to acetylsalicylate is 3.5, is aspirin a weak or strong acid? Which form is prevalent in the stomach at a pH of 1? Which form is prevalent in blood? Prove it using the Henderson-Hasselbalch equation.
Acetylsalicylic acid = Acetylsalicylate + H+, pKa =3.5

pH = pKa + Log [Acetylsalicylate]/[Acetylsalicylic acid]

In stomach, pH = 1; pKa = 3.5.

1 = 3.5 + Log [Acetylsalicylate]/[Acetylsalicylic acid]

-2.5 = Log [Acetylsalicylate]/[Acetylsalicylic acid]

2.5 = Log [Acetylsalicylic acid]/ [Acetylsalicylate]

316 = [Acetylsalicylic acid]/ [Acetylsalicylate]

In blood, pH = 7.4;

7.4 = 3.5 + Log [Acetylsalicylate]/[Acetylsalicylic acid]

3.9 = Log [Acetylsalicylate]/[Acetylsalicylic acid]

7943 = [Acetylsalicylate]/[Acetylsalicylic acid]

gyonuschot@une.edu