1 Titration of a Weak Base with a Strong Acid The same principles applied above are also applicable where we have: 1. Before addition of any acid, we have a solution of the weak base and calculation of the pH of the weak base should be performed as in previous sections. 2. After starting addition of the strong acid to the weak base, the salt of the weak base is formed. Therefore, a buffer solution results and you should consult previous lectures to find out how the pH of buffers is calculated.
Titration of a Weak Base with a Strong Acid The same principles applied above are also applicable where we have: 1. Before addition of any acid, we have a solution of the weak base and calculation of the pH of the weak base should be performed as in previous sections. - PowerPoint PPT Presentation
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Titration of a Weak Base with a Strong Acid
The same principles applied above are also applicable where we have:
1. Before addition of any acid, we have a solution of the weak base and calculation of the pH of the weak base should be performed as in previous sections.
2. After starting addition of the strong acid to the weak base, the salt of the weak base is formed. Therefore, a buffer solution results and you should consult previous lectures to find out how the pH of buffers is calculated.
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3. At the equivalence point, the amount of strong acid is exactly equivalent to the weak base and thus there will be 100% conversion of the weak base to its salt.
The problem now is to calculate the pH of the salt solution.
4. After the equivalence point, we would have a solution of the salt with excess strong acid. The
presence of the excess acid suppresses the dissociation of the salt in water and the pH of the
solution controlled by the excess acid only.
Now, let us apply the abovementioned concepts on an actual titration of a weak base with a
strong acid.
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Example
Find the pH of a 50 mL solution of 0.10 M NH3 (kb =
1.75x10-5) after addition of 0, 10, 25, 50, 60 and 100 mL of 0.10 M HCl.
Solution
1. After addition of 0 mL HCl
The solution is only 0.10 M in ammonia, therefore we have:
NH3 + H2O NH4+
+ OH-
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Kb = [NH4+][OH-]/[NH3]
1.75*10-5 = x * x / (0.10 – x)
kb is very small that we can assume that 0.10>>x. We
then have:
1.75*10-5 = x2 / 0.10
x = 1.3x10-3 M
Relative error = (1.3x10-3 /0.10) x 100 = 1.3 %
The assumption is valid, therefore:
[OH-] = 1.3x10-3 M
pOH = 2.88
pH = 11.12
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2. After addition of 10 mL HCl
A buffer will be formed from the base and its salt
Initial mmol NH3 = 0.10 x 50 = 5.0
mmol HCl added = 0.10 x 10 = 1.0
mmol NH3 left = 5.0 – 1.0 = 4.0
[NH3] = 4.0/60 M
mmol NH4+ formed = 1.0
[NH4+] = 1.0/60 M
NH3 + H2O NH4+
+ OH-
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Kb = [NH4+][OH-]/[NH3]
1.75*10-5 = (1.0/60 + x) * x / (4.0/60 – x)
kb is very small that we can assume that 1.0/60
>>x. We then have:
1.75*10-5 = 1.0/60 x / 4.0/60
x = 7.0x10-5
Relative error = (7.0x10-5 /1.0/60) x 100 = 0.42 %
The assumption is valid, therefore:
[OH-] = 7.0x10-5 M
pOH = 4.15
pH = 9.85
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3. After addition of 25 mL HCl
A buffer will be formed from the base and its salt
A buffer will be formed from the base and its salt
Initial mmol NH3 = 0.10 x 50 = 5.0
mmol HCl added = 0.10 x 50 = 2.5
mmol NH3 left = 5.0 – 25.0 = 0
This is the equivalence point
mmol NH4+ formed = 5.0, [NH4
+] = 5.0/100 = 0.05 M
NH4+ H+ + NH3
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Ka = 10-14/1.75x10-5 = 5.7x10-10
Ka = [H+][NH3]/[NH4+]
Ka = x * x / (0.05 – x)
Ka is very small. Assume 0.05 >> x
5.7*10-10 = x2/0.05
x = 5.33x10-6
Relative error = (5.33x10-6/0.05) x 100 = 0.011 %
The assumption is valid and the [H+] = 5.33x10-6 M
pH = 5.27
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5. After addition of 60 mL HCl
Initial mmol of NH3 = 0.10 x 50 = 5.0
Mmol HCl added = 0.10 x 60 = 6.0
Mmol HCl excess = 6.0 – 5.0 = 1.0
[H+] = 1.0/110 M
pH = 2.04
6. After addition of 100 mL HCl
Initial mmol of NH3 = 0.10 x 50 = 5.0
Mmol HCl added = 0.10 x 100 = 10.0
Mmol HCl excess = 10.0 – 5.0 = 5.0
[H+] = 5.0/150 M
pH = 1.48
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Titration of a Polyprotic Acid with a Strong BaseEach proton in a polyprotic acid is supposed to titrate separately. However, only those protons which satisfy the empirical relation ka1 > 104 ka2
can result in an observable break at the point of equivalence. For example, carbonic acid shows two breaks in the titration curve. Each one corresponds to a specific proton of the acid. The method of calculation of the pH is similar to that described above but initially for the first proton then the second. Each equivalence point requires a separate indicator to visualize the end point.
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There are few points to put in mind when dealing with problems of titration of polyprotic acids with strong
bases:
1. Before addition of any base, you only have the polyprotic acid solution and thus calculation of the
pH is straightforward as previously described.
2. When we start addition of base, the first proton is titrated and bicarbonate will form. A buffer solution
of carbonic acid and carbonate is formed and you should refer to the section on such calculations.
3. When all the first proton is titrated, all carbonic acid is now converted to bicarbonate (an amphoteric
protonated salt) and calculation of the pH is achieved using the appropriate root mean square
equation.
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4 .Further addition of base starts titrating the second proton thus some bicarbonate is converted to carbonate and a buffer is formed. Calculate the pH of the resulting buffer in the same way as in step 2.
5 .When enough base is added so that the titration of the second proton is complete, all bicarbonate is converted to carbonate and this is the second equivalence point. The pH is calculated for carbonate (unprotonated salt).
6 .Addition of excess base will make the solution basic where this will suppress the dissociation of carbonate. The hydrogen ion concentration is calculated from the concentration of excess hydroxide.
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Example
Find the pH of a 50 mL solution of a 0.10 M H2CO3
after addition of 0, 25, 50, 75, 100, and 150 mL of 0.10 M NaOH. Ka1=4.3x10-7 and ka2 = 4.8x10-11.
Solution
After addition of 0 mL NaOH
We only have the carbonic acid solution and the pH calculation for such types of solution was
discussed earlier and can be worked as below:
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
HCO3- H+ + CO3
2- ka2 = 4.8 x 10-11
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Since ka1 is much greater than ka2, we can neglect the
H+ from the second step and therefore we have:
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small
4.3*10-7= x2/0.10, x = 2.1x10-4
Relative error = (2.1x10-4/0.10) x 100 = 0.21%
The assumption is valid and [H+] = 2.1x10-4 M, pH = 3.68
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After addition of 25 mL NaOH
A buffer is formed from H2CO3 left and the formed
HCO3-
Initial mmol H2CO3 = 0.10 x 50 = 5.0
Mmol NaOH added = 0.10 x 25 = 2.5
Mmol H2CO3 left = 5.0 – 2.5 = 2.5
[H2CO3] = 2.5/75 M
mmol HCO3- formed = 2.5
[HCO3-] = 2.5/75 M
H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7
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ka1 = x(2.5/75 + x)/(2.5/75 – x)
ka1 is very small and in presence of the common
ion the dissociation will be further suppressed. Therefore, assume 2.5/75>>x.
x = 4.3x10-7 MRelative error = {4.3x10-7/(2.5/75)} x 100 = 0.0013%
The assumption is valid[H+] = 4.3x10-7 M
pH = 6.37
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After addition of 50 mL NaOH
Initial mmol H2CO3 = 0.10 x 50 = 5.0
mmol NaOH added = 0.10 x 50 = 5.0
mmol H2CO3 left = 5.0 – 5.0 = 0
This is the first equivalence point
mmol HCO3- formed = 5.0
[HCO3-] = 5.0/100 = 0.05 M
Now the solution contains only the protonated salt. Calculation of the pH can be done using the relation