QA Aula 3 Equilíbrio Químico e Iônico

Prévia do material em texto

Profa Dra Silvania Maria Netto
2017
QUÍMICA ANALÍTICA 
QUALITATIVA
 pH e pOH
 pH de Sais Hidrolisados
 Efeito do Íon Comum
 Solução Tampão
 Titulação Ácido-BasePARTE II
Capítulo 9 e 10
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INTRODUÇÃO
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pH e pOH
 The [H3O
+] and [OH-] in an aqueous
solution vary widely depending on the
acid or base present and its
concentration
 In general, the [H3O
+] in aqueous
solutions can range from about 10
mol/L down to about 10-15 mol/L. The
[OH] can also vary over the same
range in aqueous solution.
pH = – log [H3O
+]
and
pOH = - log [OH-] 
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pH e pOH
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pH e pOH
1. Calculate the pH of a 0.15 M solution of sodium acetate (CH3COONa).
What is the percent hydrolysis?
(kb=5.6x10
-10)
Equilibrium: NaC2H3O2(aq) ⇌ Na
+
(aq) + C2H3O2
-
(aq)

Initial [ ], M 0.15 0.00 0.00 
Change, D[ ], M -x +x +x
Equil. [ ], M (0.15 – x) x x

10-
2
3
23 10 x 5.6 
) - 15.0(
 
COONa][CH
]CO][CH[Na
 

x
x
Kb
x = 9.2x10-6 = [OH-]
pOH = - log 9.2x10-6 = 5.04
pH = 14 – 5.04  pH = 8.96
% hydrolysis = x100
9.2x10-6 M
0.15 M
% hydrolysis = 0.0061 % 
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pH e pOH
Salts that Produce Acidic Solutions
5-
2
3
233 10 x 1.8 
) - 1.00(
 
COOH][CH
]CO][CHO[H
 

x
x
Ka
-3-5 10x 4.2 )10x 1.8x 00.1( 
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pH e pOH
Salts that Produce Acidic or Basic Solutions
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pH e pOH
Salts that Produce Acidic or Basic Solutions
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pH e pOH
Salts that
Produce Acidic
or Basic
Solutions
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pH e pOH
Salts that Produce Acidic or Basic Solutions
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Some generalizations can be made about acid-base
neutralization reactions in aqueous solution and the pH of the
resulting salt solutions at 25 oC.
pH e pOH
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The common ion effect is the
shift in equilibrium caused by
the addition of a compound
having an ion in common with
the dissolved substance.
EFEITO do ÍON COMUM
The presence of a common
ion suppresses the
ionization of a weak acid or
a weak base.
Consider mixture of CH3COONa (strong
electrolyte) and CH3COOH (weak acid)
CH3COONa(s) Na
+
(aq) + CH3COO
-
(aq)
CH3COOH(aq) H
+
(aq) + CH3COO
-
(aq)
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pH of weak acid and the Common Ion Effect
2. Consider the following solutions:
a) Calculate the pH of 1.00 M HC2H3O2 (acetic acid) solution.
b) What is the pH of a solution that contains 1.00 M HC2H3O2
and 0.50 M NaC2H3O2 (sodium acetate).
EXERCÍCIO
(a):
Equilibrium: HC2H3O2(aq) ⇌ H
+
(aq) + C2H3O2
-
(aq)

Initial [ ], M 1.00 0.00 0.00 
Change, D[ ], M -x +x +x
Equil. [ ], M (1.00 – x) x x

5-
2
3
233 10 x 1.8 
) - 1.00(
 
COOH][CH
]CO][CHO[H
 

x
x
K a 2.38 pH 
M 10 x 4.2 x ]O[H
10 x 4.2 )10 x 1.8 x (1.00x
3-
3
-3-5




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(b):
Equilibrium: HC2H3O2(aq) ⇌ H
+
(aq) + C2H3O2
-
(aq)
Initial [ ], M 1.00 0.00 0.50 
Change, D[ ], M -x +x +x
Equil. [ ], M (1.00 – x) x (0.50 + x)
5-
3
233 10x 1.8 
) - 1.00(
)50.0)((
 
COOH][CH
]CO][CHO[H
 



x
xx
Ka
By approximation, 
x = (1.00/0.50)(1.8 x 10-5) = 3.6 x 10-6 M
[H+] = x = 3.6 x 10-6 M,  pH = 4.44
Solution containing the common ion is less acidic than one
containing only HC2H3O2 at the same concentration.
EXERCÍCIO
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SOLUÇÃO TAMPÃO
Solving problems with buffered solutions
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SOLUÇÃO TAMPÃO
Buffering: how does it work?
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SOLUÇÃO TAMPÃO
Buffering: how does it work?
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SOLUÇÃO TAMPÃO
Buffer solutions
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SOLUÇÃO TAMPÃO
Buffer solutions
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 Acid Ionization Constants
-
(aq)(aq)3(l)2(aq) A OH OH HA 

[HA]
]][AO[H
 
acid] [conjugate
base] e][conjugatO[H
 K 33

a
[H3O
+] = Ka x [HA]/[A
-]
-log [H3O
+] = -log Ka - log ([HA]/[A
-])
pH = pKa + log ([A
-]/[HA])
pH = pKa + log ([base]/[acid])
orHenderson-Hasselbalch 
Equation
For a particular buffer system, solutions with the same [A–]/[HA] ratio 
have same pH
SOLUÇÃO TAMPÃO
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SOLUÇÃO TAMPÃO
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SOLUÇÃO TAMPÃO
Buffer solutions
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 An important example of a buffered
system is that found in blood, which
must be maintained at a pH of 7.4
in humans.
www.chemistry.wustl.edu/~edudev/LabTutorials/Buffer/Buffer.html
SOLUÇÃO TAMPÃO
 The control of pH is often important in industrial processes.
For example, in the mashing
of barley malt, the first step
of making beer, the pH of
the solution must be
maintained at 5.0 to 5.2, so
that the protease and
peptidase enzymes can
hydrolyze the proteins from
the barley. The inventor of
the pH scale, Søren
Sørensen, was a research
scientist in a brewery.
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3. What is the pH of a buffer solution that is 0.45 M acetic
acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)?
The Ka for acetic acid is 1.8 × 10
–5.
Solution:
pH = pKa + log([C2H3O2-]/[HC2H3O2]
pH = -log(1.8 × 10–5) + log(0.85/0.45)
pH = 4.74 + 0.28
pH = 5.02
EXERCÍCIO
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Characteristics of Buffer Solutions
 Contain weak acids or weak bases and their corresponding conjugate
partners (common ions)
 Resist changes in pH
 Buffering capacity depends on concentrations of weak acid or weak
base and their common ions
 Effective pH buffering range ~ pKa  1
 Buffers contain relatively large amounts of the weak acids (HA) and
their conjugate base (A־) (or weak bases and their conjugate acids)
 Buffer pH is determined by the pKa of the acid HA and the molar ratio
of the conjugate base to acid: [A־]/[HA]
 Buffer pH changes very little because the ratio [A־]/[HA] changes very
little when a small amount of strong acid or strong base is added
SOLUÇÃO TAMPÃO
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SOLUÇÃO TAMPÃO
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Selecting an Appropriate Conjugate Acid-Base Buffer Pair
SOLUÇÃO TAMPÃO
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Addition of Acid or Base to a Buffer
SOLUÇÃO TAMPÃO
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The pH Change on Addition of Acid or Base to a Buffer
SOLUÇÃO TAMPÃO
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Some common buffers
SOLUÇÃO TAMPÃO
Buffer pka pH Range
HCHO2 – NaCHO2 3.74 2.74 – 4.74
CH3CO2H – NaCH3CO2 4.74 3.74 – 5.74
KH2PO4 – K2HPO4 7.21 6.20 – 8.20 
CO2/H2O – NaHCO3 6.37 5.40 – 7.40
NH4Cl – NH3 9.25 8.25 – 10.25
Choosing a Buffer System
 The weak acid in buffer has pKa close to target pH
 For example, KH2PO4 and K2HPO4 may be used to buffer at pH ~ 7.5
(H2PO4־ has pKa = 7.20)
 Phosphate buffer is most effective in the pH range 6.20 – 8.20; it has
the highest buffering capacity at about pH = 7.20
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4. A phosphate buffer with pH = 7.40 is prepared using KH2PO4
and K2HPO4.
a) What is the molar ratio of [HPO42-] to [H2PO4-] in the
buffered solution?
b) If [H2PO4-] = 0.20 M, what is [HPO42-]?
c) How many grams of KH2PO4 and K2HPO4, respectively, are
needed to make 500 mL of this solution? (H2PO4-
Ka=6.2x10-8)
EXERCÍCIO
a) Use Henderson-Hasselbalch equation:
pH = pKa + log([HPO42-]/[H2PO4-]) 
7.40 = 7.21 + log([HPO42-]/[H2PO4-]) 
log([HPO42-]/[H2PO4-]) = 7.40–7.21 = 0.19
[HPO42-]/[H2PO4-] = 100.19 = 1.55
(b) If [H2PO4-] = 0.20 M, 
[HPO42-] = 1.55 x 0.20 M
[HPO42-] = 0.31 M
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EXERCÍCIO
(c) Moles of KH2PO4
500 mL x (1 L/1000 mL) x 0.20 mol/L = 0.10 mol
Mass of KH2PO4
0.10 mol x (136.086 g/mol) = 14 g
Moles of K2HPO4
500 mL x (1 L/1000 mL) x 0.31 mol/L = 0.155 mol
Mass of K2HPO4
0.155 mol x (174.178 g/mol) = 27 g
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Answer: a) Yes; b) Yes; c) No; d) No; e) Yes
5. Indicate whether each of the following mixtures makes a buffer
solution. Explain.
a) 50.0 mL of 0.20 M CH3CO2H + 50.0 mL of 0.20 M NaCH3CO2b) 50.0 mL of 0.20 M HC2H3O2 + 50.0 mL of 0.10 M NaOH
c) 50.0 mL of 0.20 M HC2H3O2 + 50.0 mL of 0.20 M NaOH
d) 50.0 mL of 0.20 M NaC2H3O2 + 50.0 mL of 0.20 M HCl
e) 50.0 mL of 0.20 M NaC2H3O2 + 50.0 mL of 0.10 M HCl
EXERCÍCIO
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6. Indicate whether each of the following mixtures makes a buffer
solution. Explain.
a) 50.0 mL of 0.10 M NH3 + 50.0 mL of 0.10 M NH4NO3
b) 50.0 mL of 0.10 M NH3 + 50.0 mL of 0.10 M HNO3
c) 50.0 mL of 0.10 M NH3 + 25.0 mL of 0.10 M HNO3
d) 50.0 mL of 0.10 M NH4NO3 + 25.0 mL of 0.10 M NaOH
e) 50.0 mL of 0.10 M NH4NO3 + 50.0 mL of 0.10 M NaOH
EXERCÍCIO
Answer: a) Yes; b) No; c) Yes; d) Yes; e) No
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TITULAÇÃO ÁCIDO-BASE
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Detection of the Equivalence Point
TITULAÇÃO ÁCIDO-BASE
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TITULAÇÃO ÁCIDO-BASE
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Strong Acid – Strong Base Titrations
TITULAÇÃO ÁCIDO-BASE
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7. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100 M HCl. Calculate
the pH of the solution at these four points:
a) Before any titrant is added
b) After 40.0 mL of titrant has been added
c) After 50.0 mL of NaOH has been added. What indicator - methyl red,
bromthymol blue, or phenolphthalein - can be used to detect the
equivalence point?
d) After 50.2 mL NaOH has been added.
EXERCÍCIO
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EXERCÍCIO
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Weak Acid – Strong Base Titrations
TITULAÇÃO ÁCIDO-BASE
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8. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100 M acetic acid
(Ka=1.8x10-5). Calculate the pH of the solution at these three points:
a) After 40.0 mL of titrant has been added
b) After 50.0 mL of NaOH has been added
c) After 50.2 mL of NaOH has been added.
EXERCÍCIO
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EXERCÍCIO
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EXERCÍCIO
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Weak Base – Strong Acid Titrations
TITULAÇÃO ÁCIDO-BASE
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 Polyprotic acids - those with more than one ionizable hydrogen -react stepwise
when titrated with bases, one step for each ionizable hydrogen. If the Ka values
of the ionizable forms of the acid are sufficiently different, the titration curve has
an equivalence point for each of the ionizable hydrogens removed from the acid
molecule by titration
 For example, maleic acid, HOOC-CH=CH-COOH, is a diprotic acid with two
ionizable hydrogens, one from each -COOH group. Its titration with NaOH
occurs in two steps:
Polyprotic Acids
TITULAÇÃO ÁCIDO-BASE
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TITULAÇÃO ÁCIDO-BASE
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 The Henderson-Hasselbalch Equation
 The common ion effect tends to suppress the ionization of a weak acid or a weak
base. This action can be explained by Le Chatelier’s principle.
 A buffer solution is a combination of either a weak acid and its weak conjugate base
(supplied by a salt) or a weak base and its weak conjugate acid (supplied by a salt);
the solution reacts with small amounts of added acid or base in such a way that the
pH of the solution remains nearly constant
 The pH at the equivalence point of an acid-base titration depends on hydrolysis of
the salt formed in the neutralization reaction
 For strong acid–strong base titrations, the pH at the equivalence point is = 7
 For weak acid–strong base titrations, the pH at the equivalence point is > 7
 For strong acid–weak base titrations, the pH at the equivalence point is < 7
 Acid-base indicators are weak organic acids or bases that change color at the
equivalence point in an acid-base neutralization reaction
 The presence of a common ion decreases the solubility of a salt
 The solubility of slightly soluble salts containing basic anions increases as the
hydrogen ion concentration increases. The solubility of salts with anions derived
from strong acids is unaffected by pH.
RESUMINDO...