Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.
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Transcript of Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.
![Page 1: Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.](https://reader033.fdocuments.us/reader033/viewer/2022051400/551acb5d55034606048b4ee0/html5/thumbnails/1.jpg)
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
![Page 2: Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.](https://reader033.fdocuments.us/reader033/viewer/2022051400/551acb5d55034606048b4ee0/html5/thumbnails/2.jpg)
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
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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
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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
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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
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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][
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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
![Page 8: Chapter 16: Equilibrium in Acid-Base Systems 16.1a: Self-ionization of Water K w pH and pOH.](https://reader033.fdocuments.us/reader033/viewer/2022051400/551acb5d55034606048b4ee0/html5/thumbnails/8.jpg)
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
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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
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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
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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
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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
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HomeworkHomework
Textbook p716 #1-6Textbook p716 #1-6 p718 #7p718 #7 LSM 16.1BLSM 16.1B