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Titrations
>> Parts of this equation/concept include:
| A. Volume at the Equivalence Point (Ve) |
The equivalence point is the volume when the reactants are in exact
stoichiometric ratio. This means that you can use stoichiometry,
as described in Chapters 4
and 5, to determine the equivalence
point. No special techniques are needed.
However, if both reactants are in solution, Equation 16.25 might
be faster.
| MAVA |
= |
|
 |
MBVB |
where M is molarity, V is volume, A is acid, and
B is base. X represents the ratio of stoichiometric coefficient
of the acid to the stoichiometric coefficient of the base.
Both techniques require a balanced chemical reaction.
>> Example 1
What is the equivalence point if 30.0 mL of 0.100 M HCl
is titrated with 0.150 M NaOH?
Solution:
The reaction between NaOH and HCl is
NaOH + HCl 
NaCl + H2O
Method 1.
Method 2.
| MAVA |
= |
|
|
MBVB |
(0.100 M)(30.0 mL) = (1)(0.15 M)(VB)
20.0 mL = VB
>> Example 2
What is the equivalence point when 50.0 mL of 0.16 M NH3
is titrated with 0.25 M H2SO4?
Solution:
2 NH3 + H2SO4
(NH4)2SO4
Method 1.
Method 2.
| MAVA |
= |
|
|
MBVB |
(0.25 M)VA = (1/2)(0.16 M)(50.0
mL)
VA = 16 mL
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| B. Sketching the Titration Curve (Qualitative Titration Curves) |
The four regions of a titration curve can be predicted by considering
the titration reaction. They are
- Before any titrant is added
- Before the equivalence point
- At the equivalence point (Ve)
- After the equivalence point
Don't forget what is "obvious" overall: whenever base is added
(base is the titrant), pH will increase. Whenever acid is added
(acid is the titrant), pH will decrease.
In region 1, the pH is determined from the reactant alone. This
process was described earlier for weak acids (using Ka),
weak bases (using Kb), and strong acids and bases
(using stoichiometry).
In region 2, some titrant has been added. The titrant will completely
react with the reactant being titrated. Therefore the concentration
of reactant will decrease and the concentration of its conjugate
acid (or base) will increase.
If the reactant is a strong acid or base, its conjugate is too
weak to react with water (neutral) and will not affect pH. Thus
a slow but steady change in pH is observed.
If the reactant is a weak acid or base, a buffer solution is formed
and the area before the equivalence point is nearly horizontal.
The most useful point is halfway to the equivalence point. Since
half of the reactant has reacted, forming its conjugate, the concentration
of acid and base are equal. At that volume, pH = pKa.
In region 3, all the reactant has been converted to its conjugate.
The conjugate has also been diluted. The concentration of the conjugate
acid or base can be determined from MRVR/VT,
where MR = molarity of reactant, VR
= volume of reactant and VT = total volume (VR
+ Ve). This molarity is then used to calculate
pH with the Ka or Kb of the
conjugate of the original reactant.
In region 4, there is an excess of titrant. Because the titrant
is a strong acid or base, it will control pH. Here, too, dilution
must be taken into account. The concentration of the titrant in
the reactant solution can be determined similarly to that for the
conjugate:
| (M of titrant)(V of titrant
beyond the equivalence point) |
|
| (VR + Vtotal
titrant added) |
|
= |
molarity |
>> Example 3
Sketch the titration curve for the titration of 25.0 mL of 0.10
M HCl with 0.10 M NaOH.
Solution:
HCl is the reactant; NaOH is the titrant. The reaction is
HCl + NaOH
H2O + NaCl
The phrase "sketch" implies a qualitative answer.
Region 1. At volume zero, before any titrant has been added.
HCl is present in the reaction solution. It is a strong acid,
so the pH will be very low.
Region 2. HCl is still present, so the pH is still very low,
but higher than the starting point.
Region 3. HCl is all used up; all the NaOH added has also reacted.
The only substances present are water and sodium chloride. Since
NaCl does not affect pH, the pH is the pH of water, 7.
Region 4. The NaOH added does not have anything to react with,
so the pH is characteristic of a strong base, very high. It increases
as more NaOH is added.
>> Example 4
Sketch the titration curve for the titration of 30.0 mL of 0.10
M HF with 0.20 M KOH.
Solution:
HF is the reactant; KOH is the titrant. The reaction is
HF + KOH
KF + H2O (or K+ + F– +
H2O)
Region 1. At volume zero, only HF is present. It is a weak acid,
so the pH will be less than 7.00 but higher than what it would
be for a strong acid.
Region 2. Some of the HF has reacted with the KOH, making the
conjugate base F–. The presence of both acid and
base creates a buffer solution. Thus the pH in this region is
higher than that in region 1, but relatively constant (slightly
increasing) and centered around 3.45, the pKa
of HF.
Region 3. All the HF is used up and all the KOH has reacted,
but F– is present. F– is a weak
base, so the pH will be slightly greater than 7.
Region 4. F– is still present, but now OH–
is present too. Consequently, pH will be controlled by the strong
base and will be very high.
>> Example 5
Sketch the titration curve for 20.0 mL of 0.15 M NH3 with 0.10 M HCl.
Solution:
NH3 is the reactant; HCl is the titrant. The reaction
is
NH3 + HCl
NH4+ + Cl–
Region 1. Only NH3 is present. It is a weak base,
so the pH will be somewhat more than 7.
Region 2. Some NH3 has reacted, but some remains.
Some NH4+ has been created. Consequently,
there is a buffer solution. The pH is decreasing, but slightly.
The pH is near the pKa of NH4+
(the conjugate acid) = 9.25.
Region 3. Only NH4+ is present. It is a
weak acid, so the pH is slightly less than 7.
Region 4. Not only weak acid, NH4+, but
strong acid HCl are present. Consequently, the pH is very low.
>> back
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| C. Quantitative Titration Curves |
>> Example 6
What is the pH at 0.0, 20.0, and 30.0 mL in the titration of
25.0 mL of 0.10 M HCl with 0.10 M NaOH.
Solution:
This is the same titration described in Example 3. Using MAVA
= (1/X)MBVB, where
X = 1, the volume at the equivalence point is 25.0 mL.
HCl is the reactant; NaOH is the titrant. The reaction is
HCl + NaOH
H2O + NaCl
0.0 mL is region 1. 0.10 M HCl is present. Since HCl is
a strong acid, the concentration of [H3O+]
= 0.10 M. Therefore pH = 1.00.
20.0 mL is before the equivalence point (region 2). The moles
of acid present originally are
| 0.025 L |
|
|
= |
0.0025 mol HCl to start |
The moles of base added are
| 0.020 L |
|
|
= |
0.0020 mol NaOH |
The moles of acid used up are
| 0.0020 mol NaOH |
|
|
= |
0.0020 mol HCl used up |
The moles of acid remaining are
0.0025 – 0.0020 = 0.0005 mol left over
The concentration of HCl is now
| 0.0005 mol HCl |
|
| 0.025 + 0.020 |
|
= |
0.011 M |
So the pH is 1.95.
30 mL is 5.0 mL after the equivalence point. The moles of NaOH
in the reaction solution are
| 0.0050 L |
|
|
= |
0.00050 mol NaOH |
It is in a total volume of 25.0 mL + 30.0 mL = 55.0 mL or 0.0550
L
pOH = 2.04
pH = 11.96
>> Example 7
In the titration of 30.0 mL of 0.10 M HF with 0.20 M
KOH, what is the pH at 10.0, 15.0, and 20.0 mL?
Solution:
This is the same titration as in Example 4. Using MAVA
= (1/X)MBVB, where
X = 1, the volume at the equivalence point is 15.0 mL.
10.0 mL is in region 2, before the equivalence point. The initial
moles of HF are
| 0.0300 L |
|
|
= |
0.0030 mol HF |
The moles of KOH added are
| 0.010 L |
|
|
= |
0.0020 mol KOH |
The moles of F– created are
| 0.0020 mol KOH |
|
|
= |
0.0020 mol F– made (base) |
The moles of acid reacted are
| 0.0020 mol KOH |
|
|
= |
0.0020 mol HF used up |
The moles of acid left over are
0.0030 mol – 0.0020 mol = 0.0010 mol HF left (acid)
Since this is a buffer, the Henderson–Hasselbalch equation
can be used. Because both acid and conjugate base are in the same
volume of solution, the volume cancels in the acid–base ratio
and the mole ratio can be used directly.
pH = 3.46 + 0.30 = 3.76
15.0 mL is region 3, the equivalence point. Only F–
is present. Since all the HF has been converted to F–,
there is 0.0030 mol F–. The total volume is 15.0
mL + 30.0 mL = 45.0 mL = 0.045 L. Consequently, [F–]
= 0.067 M.
Since F– is a weak base, its reaction with water
is
F– + H2O 
HF + OH–
Using this equation and the ICE table, concentration of OH–
is 1.4 x 10–6. So pOH = 5.86 and pH = 8.14.
At 20.0 mL, which is 5.0 mL after the equivalence point. Consequently,
the moles of unreacted base are
| 0.0050 L |
|
|
= |
0.0010 mol |
The total volume is 20.0 mL + 30.0 mL = 50.0 mL = 0.0500 L solution.
So, [OH–] = 0.020 M, pOH = 1.70 and pH
= 12.30
>> Example 8
In the titration of 20.0 mL of 0.15 M NH3 with
0.10 M HCl, what is the pH at 10.0, 20.0, 30.0 mL?
Solution:
This is the same titration as that in Example 5.
Using MAVA = (1/X)MBVB,
where X = 1, the volume at the equivalence point is 40.0
mL
10.0 mL is region 2, before the equivalence point. The moles
of base initially are
| 0.020 L |
|
|
= |
0.0030 mol NH3 to start |
The moles of acid added are
| 0.010 L |
|
|
= |
0.0010 mol HCl added |
The moles of conjugate acid created
| 0.0010 mol HCl |
|
|
= |
0.0010 mol NH4+
created (acid) |
The moles of base used up
| 0.0010 mol HCl |
|
|
= |
0.0010 mol NH3 used up |
The moles of base remaining
0.0030 mol – 0.0010 mol = 0.0020 mol NH3 remaining
(base)
Using Henderson–Hasselbalch equation with moles instead
of molarity
The pKa of NH4+:
| Ka |
= |
|
= |
|
= |
5.6 x 10–10,
so pKa = 9.25 |
20.0 mL is region 2, halfway to the equivalence point. At halfway,
[acid] = [base], so pH = pKa. The pH is 9.25.
30.0 mL is r egion 3, the equivalence point, only NH4+
is present. There were 0.0030 mol NH4+ created
(all the moles of NH3 became NH4+).
The total volume is 30.0 mL + 20.0 mL = 50.0 mL. Therefore the
concentration of NH4+ is 0.0030 mol/0.0500
L = 0.060 M.
Its reaction with water is
NH4+ + H2O
NH3 + H3O+
Solving the Ka expression using [NH4+]
= 0.060 M and the ICE table, [H3O+]
= 5.8 x 10–6 M and pH = 5.24.
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Indicators are simply weak acids or bases where the acid and base
forms are different colors. The relative concentrations of acid
and base depend on the pKa of the indicator and
the pH. At pH values about 1 pH unit higher than the pKa,
the basic form and color predominate. It doesn't matter how much
higher; it will be the basic color. Similarly, the acid color will
be seen at any pH lower than 1 pH unit below the pKa.
Since the pH change is dramatic at the equivalence point, indicators
are chosen so that the pH of the equivalence point is near the pKa
of the indicator. The color change is how the equivalence point
is indicated.
>> Example 9
Refer to the titration in Example 7. The indicator phenol red
has a pKa of about 7.5. Its acid color is yellow;
its base color is red. What color will the indicator be at 10,
15, and 20 mL? Would this be a good indicator for this titration?
Solution:
At 10 mL, the pH is 3.76. This is very acidic; the color will
be yellow.
At 15 mL, the pH is 8.14. This is just within the ±1 of
the pKa, so it will be orange. Likely a reddish–orange.
At 20 mL, the pH is 12.30. This is very basic; the color will
be red.
Since it changes color at the equivalence point (it doesn't have
to hit exactly), this would be a reasonable choice.
>> Example 10
Refer to the titration in Example 7. The indicator methyl orange
has a pKa of about 3.5. It is red in its acidic
form and yellow in its basic form. What color will the indicator
be at 10, 15, and 20 mL? Would this be a good indicator for this
titration?
Solution:
At 10 mL, the pH is 3.76. This is near its pKa,
so the color is orange.
At 15 mL, the pH is 8.14. This far above the pKa,
so the color is yellow.
At 20 mL, the pH is 12.30. This is even more above the pKa,
so the color is still yellow. (No difference between 20 and 15
mL.)
This would be a poor choice of an indicator, since the color
change is well before the equivalence point.
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