WATER

1

• WATER
• THE HYDRONIUM ION
• AUTO-IONIZATION
• THE pH SCALE

2

• Fredrich Kohlrausch, around
• 1900, found that no matter how
• pure water is, it still conducts a
• minute amount of electric
• current. This proves that water
• self-ionizes.

3

• Since the water molecule is
• amphoteric, it may dissociate
• with itself to a slight extent.
• Only about 2 out of a billion
• water molecules are ionized at
• any instant!

4
H2O(l) H2O(l) ltgt H3O(aq) OH-(aq)

• The equilibrium expression used
• here is referred to as the Kw
• (ionization constant for water).

5

• In pure water or dilute aqueous
• solutions, the concentration of water
• can be considered to be a constant
• (55.4 M), so we include that with the
• equilibrium constant and write the
• expression as
• KeqH2O2 Kw H3OOH-

6

• Kw 1.0 x 10-14
• (Kw 1.008 x 10-14 _at_ 25 Celsius)
• Knowing this value allows us to
• calculate the OH- and H
• concentration for various situations.

7

• OH- H solution is neutral (in
• pure water, each of these is 1.0 x 10-7)
• OH- gt H solution is basic
• OH- lt H solution is acidic

8
Kw Ka x Kb another very beneficial equation

• pH - log H
• pOH - log OH-
• pH pOH 14
• pH 6.9 and lower (acidic)
• 7.0 (neutral)
  
• 7.1 and greater (basic)

9
The pH Scale
Used to designate the H in most aqueous
solutions where H is small.
10

• Use as many decimal places as
• there are sig.figs. in the problem!
• The negative base 10 logarithm of
• the hydronium ion concentration
• becomes the whole number
• therefore, only the decimals to the
• right are significant.

11
Exercise 6 Calculating H and OH-

• Calculate H or OH- as required for
• each of the following solutions at
• 25C, and state whether the solution
• is neutral, acidic, or basic.
• a. 1.0 X 10-5 M OH-
• b. 1.0 X 10-7 M OH-
• c. 10.0 M H

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Solution

• A H 1.0 X 10-9 M, basic
• B H 1.0 X 10-7 M, neutral
• C OH- 1.0 X 10-15 M, acidic

13
Exercise 7 Calculating pH and pOH

• Calculate pH and pOH for each of
• the following solutions at 25C.
• a. 1.0 X 10-3 M OH-
• b. 1.0 M H

14
Solution

• A pH 11.00
• pOH 3.00
• B pH 0.00
• pOH 14.00

15
Exercise 8 Calculating pH

• The pH of a sample of human blood
• was measured to be 7.41 at 25C.
• Calculate pOH, H, and OH- for
• the sample.

16
Solution

• pOH 6.59
• H 3.9 X 10-8
• OH- 2.6 X 10-7 M

17
Exercise 9 pH of Strong Acids

• Calculate the pH of
• a. 0.10 M HNO3
• b. 1.0 X 10-10 M HCl

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Solution

• A pH 1.00
• B pH 7.00

19
Exercise 10 The pH of Strong Bases

• Calculate the pH of a 5.0 X 10-2 M
• NaOH solution.

20
Solution

• pH 12.70

21
Calculating pH of Weak Acid Solutions

• Calculating pH of weak acids
• involves setting up an equilibrium.

22
Always start by

• 1) writing the equation
• 2) setting up the acid equilibrium
• expression (Ka)
• 3) defining initial concentrations, changes, and
  final concentrations in terms of X

23

• 4) substituting values and variables
• into the Ka expression
• 5) solving for X
• (use the RICE diagram learned in
• general equilibrium!)

24
Calculate the pH of a 1.00 x 10-4 M solution of
acetic acid.

• The Ka of acetic acid is 1.8 x 10-5
• HC2H3O2 ? H C2H3O2-
• Ka HC2H3O2- 1.8 x 10-5
• HC2H3O2

25

• Reaction HC2H3O2 ? H C2H3O2-
• Initial 1.00 x 10-4 0
  0
• Change -x x
  x
• Equilibrium 1.00 x 10-4 - x x x

26
Often, the denominator -x in a Ka expression can
be treated as negligible. If it is 10-4 or less

• 1.8 x 10-5 (x)(x) _
• 1.00x10-4 - x
• 
  
• 1.8 x 10-5 ? (x)(x) _
• 1.00 x 10-4
• x 4.2 x 10-5

27

• When you assume that x is
• negligible, you must check the
• validity of this assumption.

28

• To be valid, x must be less than
• 5 of the number that it was to be
• subtracted from.
• dissociation "x" x 100
• original

29

• In this example, 4.2 x 10-5 is greater
• than 5 of 1.00 x 10-4.
• This means that the assumption that
• x was negligible was invalid and x
• must be solved for using the
• quadratic equation or the method of
• successive approximation.

30
Use of the Quadratic Equation
31
ax2 bx c 0
32
Using the values a 1, b 1.8x10-5, c
-1.8x10-9
and
33
Since a concentration can not be negative

• x 3.5 x 10-5 M
• x H 3.5 x 10-5
• pH -log 3.5 x 10-5 4.46

34

• Another method which some people
• prefer is the method of successive
• approximations. In this method, you
• start out assuming that x is
• negligible, solve for x, and repeatedly
• plug your value of x into the
• equation again until you get the
• same value of x two successive times.