WATER STRUCTURE

1. BIOCHEMISTRY
Worksheet #2
Name_________________________ Course/Year/Block___________ Score: ___________
WATER: Acidic and Basic Properties of Water
OBJECTIVES
1. Describe how pure water becomes ionized and identify the ions formed.
2. Define operationally the equilibrium constant for the ionization of water.
3. Calculate the ion concentration in an aqueous solution.
4. Determine logarithm and antilogarithm values.
EXPLORE
Below is a tutorial/interactive example of the concepts. Use them as guide
for answering the questions.
Absolutely pure water has no acid or base added but still contains a very small
number of [H3O+] and [OH-] ions. These ions are formed by the transfer of a proton from one
molecule of water (the proton donor) to another water molecule (the proton acceptor).
H2O + H2O  H3O+
+ OH-
The ion product of water, Kw, also called the water constant or the equilibrium
constant for the ionization of water, is equal to 1.0 x 10-14.
Kw = [H3O+
] [OH-
] = 1.0 x 10-14
This equation is important because it also applies to any water solution. When
water is pure, H3O+ and OH- form in equal amounts and concentration at 1.0 x 10-7 mol/L each.
Since there is a very small concentration, pure water does not conduct electricity and is
neutral.
Suppose 0.010 mol of HCL was added to 1 L of pure water, it reacts
completely to give H30+ and CI- ions. The concentration of H30+ will be 0.010 M, or 1.0 x 10-2 M.
This means that [0H-] must be 1.0 x 10-14 / 1.0x10-2 = 1.0x10-12 M
Example:
The [0H-] of an aqueous solution is 1.0 x 10-4 M. What is its [H30+]?
Procedure:
Determine the hydrogen ion concentration [H30+] by dividing the ion product of water,
1.0 x 10-14, to the given [0H-].
Solution:
1.0 x 10-14 = [H30+] [0H-]
[H30+] = 1.0 x 10-14
--------------------
1.0 x 10-4 M = 1.0 x 10-10
M
Did you notice that when the [0H-] concentration was high (1.0 x 10-4 M), the [H30+]
concentration was low (1.0 x 10-10 M)?
Aqueous solution can have a very high [H30+], but the [0H-] must be very low and vice
versa.
If the concentration of [H30+] in a solution is greater than 1.0 x 10-7 M, it is acidic and
should have a [0H-] concentration lower than 1.0 x 10-7 M. The same concept applies for [0H-].
CONSTRUCT YOUR LEARNING
Answer the following questions:
1. How can pure water be ionized?
2. Write the name and symbols for the ions in pure water.
3. Define operationally the ion product of water.
4. Is pure water neutral? Why do you say so?
5. Why do you think should the concentration of [H30+] need to be low when the [0H-]
concentration is high?

2. APPLY YOUR LEARNING
A. Given the following values of [H30+], calculate the corresponding value of [0H-] for each
solution:
1. 1.0 x 10-11 M
2. 1.0 x 10-7 M
3. 10 M
B. given the following values of [0H-], calculate the corresponding value of [H30+] for each of
solution:
1. 1.0 x 10-10 M
2. 1,0 x 10-2 M
3. 10 M
EXPLORE PART II
Below is a tutorial on using Logs and Antilogs. Study the tutorial and answer the
questions that follow.
We often need to use common or base 10 logarithms (logs) when dealing with acids,
bases and buffers. You can handle logs and antilogs easily.
A common logarithm is the power to which you raise 10 to get another number. The
log of 100 is 2 because you raise 10 the second power to get 10. Similarly, the log of 1000 is 3
because 103 is 1000. The log of 0.1 is -1 because 10-1 is 0.1. The log can be determined
using a calculator by entering the number then pressing log or vice versa. For example, log 4.5
= 0.65.
An antilog is the reverse of a log and is also called the inverse log. If you take 10 and
raise it to a power, you are taking an antilog. The antilog of 5 is 100 000.
In the calculator, enter 5, then press INV or 2nd function, then press log, or press INV
first then log then the number.
Antilog differs from –log. Antilog 3 means that we take 10 and raise it to the power of 3
and the answer is 1000. –log 3 means that we take the log of 3, which is equal to 0.477 and
take the negative of it, thus, -0.48.
In calculating pH, you will use the negative log. pH equals –log[H+]. If you know that
[H+] is 0.01 M, to find the pH, enter it into your calculator, and press log. The answer would be
-2. Then, take the negative of that value to give a pH of 2.
CONSTRUCT YOUR LEARNING PART II
Answer the following questions:
1. Differentiate among log, antilog and – log. Give example each to elaborate.
APPLY YOUR LEARNING PARTT II
A. Using your calculator, determine the following:
1. Log of 1 = ____________________
2. Antilog of 0.25 = ____________________
3. -log of 2 = ____________________
4. -log of 1.5 x 10-10 = ____________________