Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.

Chapter 16: Chapter 16: Equilibrium in Acid-Base SystemsEquilibrium in Acid-Base Systems

16.1a: Self-ionization of Water16.1a: Self-ionization of Water

KKww

pH and pOHpH and pOH

Self-ionization of waterSelf-ionization of water

If water is a weak electrolyte, what If water is a weak electrolyte, what does it ionize in to?does it ionize in to?

HH22O O (l) (l) ⇄⇄ OH OH- - (aq) (aq) + H+ H+ +

(aq)(aq)

OROR

2H2H22O O (l) (l) ⇄⇄ OH OH- - (aq) (aq) + H+ H33OO+ +

(aq)(aq)

the concentrations of the ions are the concentrations of the ions are only 1.0 x 10only 1.0 x 10-7-7 M at 25°C M at 25°C

the product of these concentrations the product of these concentrations will remain constant for any solution will remain constant for any solution at the same temperatureat the same temperature

Self-ionization of WaterSelf-ionization of Water

concentrations or molarities can be concentrations or molarities can be written with bracketswritten with brackets

For example:For example:

concentration of A = [A] = 2.0 Mconcentration of A = [A] = 2.0 M KKww: :

the ionization constant of waterthe ionization constant of water the product of [OHthe product of [OH--] and [H] and [H++]]

at 25at 25ooCC14

3 100.1]][[ OHOHKw

KKww subject to the same restricition as any subject to the same restricition as any

other equilibrium constant (T, P)other equilibrium constant (T, P) Will acidic solutions have more HWill acidic solutions have more H++ or or

OHOH--?? [H[H++]>[OH]>[OH--]: acidic]: acidic [OH[OH--]>[H]>[H++]: basic]: basic [OH[OH--]=[H]=[H++]: neutral]: neutral

can find the [OHcan find the [OH--] or [H] or [H++] from a mole ] from a mole ratio of the dissociation or reaction in ratio of the dissociation or reaction in the water of the acid or basethe water of the acid or base

Example 1Example 1 Calculate the HCalculate the H++ concentration in a 1.5 M concentration in a 1.5 M

Ca(OH)Ca(OH)22 solution solution What does the Ca(OH)What does the Ca(OH)22 create in create in

solution?solution? Ca(OH)Ca(OH)22 Ca Ca2+2+ + 2OH + 2OH--

can calculate the OHcan calculate the OH-- concentration concentration use Kuse Kww to calculate the H to calculate the H++ concentration concentration

L

molOH

OHmolCa

molOH

L

OHmolCa

1

0.3

)(1

2

1

)(5.1

2

2

-14[ ] 3.0 M and [ ][ ] 1.0 10OH OH H

-14151.0 10

[ ] 3.3 103.0

H M

Example 2Example 2 Calculate the HCalculate the H++ and OH and OH-- concentration of concentration of

a 1.0x10a 1.0x10-4-4 M solution of HNO M solution of HNO33

Find the [HFind the [H++] from mole ratio] from mole ratio Find [OHFind [OH--] from K] from Kww and [H and [H++]]

HNOHNO33 H H++ + NO + NO33--

L

molH

molHNO

molH

L

molHNO

1

100.1

1

1

1

100.1 4

3

34

-4 -14[ ] 1.0 10 M and [ ][ ] 1.0 10H OH H

MOH 104

-14

100.1100.1

101.0][

pH scalepH scale more convenient than more convenient than

using concentrationsusing concentrations pH=-log [HpH=-log [H++]] pOH=-log [OHpOH=-log [OH--]] pH increases as [HpH increases as [H++] ]

decreasesdecreases pH < 7: acidpH < 7: acid pH > 7: basepH > 7: base pH = 7: neutralpH = 7: neutral

pHpH for any solution at 25for any solution at 25ooC:C:

-log[H-log[H++] + -log[OH] + -log[OH--] = -log([H] = -log([H++][OH][OH--] ] = -log(1.0x10= -log(1.0x10-14-14) =14) =14

14.00 = pH + pOH14.00 = pH + pOH

Example 3Example 3

Find the pH of a 1.0x10Find the pH of a 1.0x10-3-3 M NaOH M NaOH solutionsolution find [OHfind [OH--] using mole ratio] using mole ratio find [Hfind [H++] from K] from Kww

find pH from [Hfind pH from [H++]]

OROR find pOH from [OHfind pOH from [OH--] ] find pH from pOHfind pH from pOH

Example 3 sol’nExample 3 sol’n

00.11)100.1log(]log[ 11 HpH

L

molOH

molNaOH

molOH

L

molNaOH

1

100.1

1

1

1

100.1 33

MH 113

-143 100.1

100.1

101.0][ M 100.1][OH

OR

00.1100.14

00.3)100.1log(]log[ 3

pOHpH

OHpOH

Calculating [HCalculating [H++] and [OH] and [OH--] ]

reverse the pH equationreverse the pH equation

The pH of a solution is 7.52. Find the The pH of a solution is 7.52. Find the [H+] and [OH-] and determine whether it [H+] and [OH-] and determine whether it is acidic, basic, or neutral.is acidic, basic, or neutral.

pH pOH[H ] 10 and [OH ] 10

7.52 -8

(14.00 7.52) 7

[H ] 10 3.0 10

[OH ] 10 3.3 10

[OH-] > [H+] so solution is basic

ExampleExample

A shampoo has a pH of 2.53. A shampoo has a pH of 2.53. Calculate the pOH, [HCalculate the pOH, [H++] and [OH] and [OH--]. Is ]. Is it acidic, basic, or neutral?it acidic, basic, or neutral?

14.00 14.00 2.53 11.47pOH pH

12

2.53

11.47

[ ] 10 0.0029

[OH ] 10 3.42 10

H M

M

pH < 7 so acidic

HomeworkHomework

Textbook p716 #1-6Textbook p716 #1-6 p718 #7p718 #7 LSM 16.1BLSM 16.1B