Lecture 54 Acids and Bases IV

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Lecture 54 - Acids and Bases IV
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Autodissociation of water can be significant...

• e.g. 1.0 x 10-7 M HNO3(aq)
• find pH !
• H3O(aq) from H2O H3O(aq) from HNO3
• but H3O ¹ 2 x 10-7
• 2 H2O(l) ¾ H3O(aq) OH-(aq)

equilibrium shift due to H3O from HNO3
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Autodissociation of water can be significant...

• solution
• positive charge negative charge
• i.e. H3O (NO3- OH-)
• but, Kw H3O OH-

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Kw
thus, H3O NO3-
H3O
thus, H3O2 NO3- H3O Kw
thus, H3O2 - NO3- H3O - Kw 0
since NO3- 1 x 10-7, H3O 1.6 x 10-7 M
pH 6.80
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Acid Base Indicators

• are weak acids or bases themselves
• Hin(aq) H2O(l) ¾ H3O(aq) In-(aq)

Color A
Color B
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Acid Base Indicators
i.e. ratio of colors changes with pH
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e.g. Alizarin yellow
H
CO2H
N - N
O
(RED)
NO2
OH-
CO2-
N N -
-OH
(YELLOW)
NO2
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Colors from Plants
Anthocyanins
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The Common Ion Effect

• HF(aq) H2O(l) ¾ H3O(aq) F-(aq)
• NaF(s) H2O(l) ? Na(aq) F-(aq)
• F-(aq) is the common ion

equilibrium shifts, pH rises
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The Common Ion Effect

• works for bases also
• NH3(aq) H2O(l) ¾ NH4(aq) OH-(aq)
• adding NH4Cl(s)
• NH4Cl(s) H2O ? NH4(aq) Cl-(aq)

equilibrium shifts, pH falls
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The common ion effect

• e.g. 2.0 M HF(aq) had pH 1.4
• Add NaF(s) until NaF(aq) 2.0 M, find pH
• (Ka (HF) 7.2 x 10-4)

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The common ion effect

• HF(aq) H2O(l) ¾ H3O(aq) F-(aq)
• NaF(s) H2O(l) ? Na(aq) F-(aq)
• Initial Equilibrium
• HF(aq) 2.0 2.0 - x
• F- 2.0 2.0 x (from NaF HF)
• H3O 0 x (from HF dissociation)

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The common ion effect
x (2.0 x)

2.0 - x
x(2.0)

(x is small)
2.0
x
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The common ion effect

• H3O x 7.2 x 10-4 M
• pH -log10 (7.2 x 10-4)
• 3.14
• (was 1.4 before addition of NaF)

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Application of the common ion effect - buffered
solutions

• buffered solutions resist changes in pH
• e.g. blood pH is very constant at 7.4

a weak acid and its salt or a weak base
and its salt
a buffer
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An unbuffered solution...

• 1.0 L H2O, add 0.05 mol NaOH, find pH
• pH 14 - pOH
• 14 - (-log10(0.05))
• 12.7
• (was 7.00 before adding base)

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Buffered Solutions

• e.g. 1.0 L of 2.0 M HF(aq), 2.0 M NaF, pH 3.14
• now add 0.05 mol NaOH(s), find new pH

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strategy...

• 1. react base with acid, some acid is destroyed.
• 2. residual acid re-equilibrates, find new pH

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1. React acid with base

• HF(aq) OH- H2O F-

1 L x 2 M 2 mol
1 L x 2 M 2 mol
0.05 mol
initial
2.0 .05 2.05 mol
2 - 0.05 1.95 mol
0.0
final
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2. Acid re-equilibrates

• HF H2O ¾ H3O F-

0
2.05
initial
1.95
x
2.05x
1.95-x
final
x(2.05)

1.95
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2. Acid re-equilibrates
x(2.05)
x (2.05x)
7.2 x 10-4

1.95 - x
1.95
solving, x H3O 6.85 x 10-4
pH 3.16
(was 3.14 before adding base)