Chemistry 201: General Chemistry II - Lecture
Dr. Namphol Sinkaset
Chapter 18 Study Guide
Concepts
1. A buffer is a solution that resists changes in pH by neutralizing added acid or base.
2. Buffers are created using conjugate acid/base pairs.
3. The pH of a buffer can be calculated by approaching the system as an equilibrium
problem. Since a conjugate acid/base pair are both in solution, the problem can be
solved from a Ka or Kb point of view.
4. The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer
or the amounts/concentrations necessary to create a buffer. It works provided the “x
is small” approximation is valid.
5. Adding strong acid or strong base will result in small changes in the pH of a buffer.
New concentrations of the conjugate acid/base pair must calculated based on the stoichiometric reaction and the new total volume.
6. Note that in the Henderson-Hasselbalch equation, the ratio of base to acid is what’s
needed. As such, moles can be used in place of concentrations since the volume of the
entire buffer is the same for both base and acid.
7. Buffers can be made with a weak base and its conjugate acid as well. In these problems,
the pKa of the conjugate acid must be used in the Henderson-Hasselbalch equation.
8. Effective buffers have relative concentrations of conjugate acid/base that do not differ
by more than a factor of 10 and have high actual concentrations of conjugate acid and
base. The effective range for a buffer system is ±1 pH unit on either side of pKa .
9. Buffer capacity is the amount of acid or base that can be added to a buffer without
destroying its effectiveness.
10. A buffer is destroyed when either of the conjugate acid/base pairs has been used up.
Thus, higher concentrations of buffer components lead to higher buffer capacity.
11. In an acid-base titration, a basic (or acidic) solution of unknown concentration is
reacted with an acidic (or basic) solution of known concentration.
12. An indicator is a substance used to visualize the endpoint of a titration.
13. At the equivalence point, the number of moles of acid equals the number of moles
of base.
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14. Theoretical titration curves (also known as pH curves) can be plotted by calculating
the pH along several points of the titration.
15. The two keys to titration curve calculations are writing the titration equation and
finding the equivalence point of the titration.
16. Regardless of the type of titration (strong/strong or weak/strong), the titration curve
can be broken up into four regions: (1) before the titration begins; (2) pre-equivalence;
(3) at equivalence; and (4) post-equivalence. The type of calculation performed depends on the region!
17. For weak/strong titrations, the pH does not equal 7.00 at the equivalence point. Additionally, at half equivalence, pH = pKa .
18. Polyprotic acids will display multiple equivalence points in their titration curves if the
Ka ’s are different enough. The volumes between equivalence points will always be
equal.
19. An indicator is itself a weak acid that has a different color than its conjugate base.
20. If pH > pKa of HIn, then the color of the solution will be that of In− . If pH = pKa
of HIn, then the color of the solution will be in between that of HIn and In− . If pH <
pKa of HIn, then the color of the solution will be that of HIn.
21. Solubility is not as clear cut as it seems; there are different degrees of solubility.
22. The solubility product constant, Ksp , is an equilibrium constant that describes the
relationship between the concentrations of the constituent ions and their corresponding
precipitate. Just like any other equilibrium constant, it tells you how far the reaction
goes towards products.
23. Generally, solubility is defined as the amount of grams of a solid that will dissolve in
100 g of water. Molar solubility is an alternative measure of solubility; obviously, it
is the number of moles of a solid that can dissolve per L of solution.
24. Ksp ’s are used to calculate molar solubilities, and these problems are treated like any
other equilibrium problem.
25. common ion effect: the lowering of the solubility of an ionic compound by the
addition of a common ion (i.e., an ion involved in the Ksp expression).
26. pH will affect the solubility of an ionic compound if either H3 O+ or OH− can react
with the cation or anion.
27. Q calculations can be performed to see whether or not a precipitate will form.
28. If Q > Ksp , a precipitate will form.
29. If Q = Ksp , no precipitate will form, but the solution is saturated.
30. If Q < Ksp , no precipitate will form.
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31. A complex ion is comprised of a transition metal Lewis acid that is bound to one or
more Lewis bases. In a complex ion, the Lewis bases attached to the central metal are
called ligands.
32. Formation of complex ions can also be written as an equilibrium, and the equilibrium
constant in this case is called a formation constant, Kf .
33. Formation constants are usually very large.
34. Calculations with Kf ’s are a little different as we assume that the equilibrium lies all
the way to the right.
35. Formation of complex ions will enhance the solubility of otherwise insoluble ionic compounds. Thus, the addition of Lewis bases that can bind to metals will increase solubility.
36. Some metal hydroxides are amphoteric. Thus, their solubility is very pH dependent.
37. Only aluminum, chromium(III), zinc(II), lead(II), and tin(II) hydroxides are amphoteric.
Equations
1. pH = pKa + log
[base]i
(Henderson-Hasselbalch equation)
[acid]i
2. pKa + pKb = 14.00 (Relationship between pKa and pKb )
−
3. Ksp = [A+ ][B− ] (solubility product constant for AB(s) *
) A+
(aq) + B(aq) )
4. Kf =
[MLx+
y ]
x+
*
(formation constant for Mx+
x+
(aq) + yL(aq) ) MLy(aq) )
[M ][L]y
Representative Problems
R7. How many grams of sodium formate, NaCHO2 , would have to be dissolved
in 1.0 L of 0.12 M formic acid (pKa 3.74) to make the solution a buffer for pH
3.80?
Since we know it’s a buffer problem, we know we will need to use the HendersonHasselbalch equation.
[A− ]
[HA]
[CHO−
2]
pH = pKa + log
[CHO2 H]
[CHO−
2]
3.80 = 3.74 + log
[0.12]
pH = pKa + log
3
[CHO−
2]
[0.12]
[CHO−
2]
1.148 =
[0.12]
−
[CHO2 ] = 0.13776
0.06 = log
We need to do a stoichiometry calculation to get the grams of sodium formate needed.
1.0 L ×
1 mole NaCHO2 68.008 g NaCHO2
0.13776 mole CHO−
2
×
= 9.37 g NaCHO2
×
1L
1 mole CHO−
1 mole NaCHO2
2
R22. Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr.
Find the pH at the following volumes of acid added: Va = 5.0, Veq , and 12.0 mL.
As in all titrations, there are two things you should always do first. One is write the
titration reaction, and the other is find the equivalence point. The titration is between a
strong base and a strong acid. Therefore, the reaction is:
+
OH−
(aq) + H3 O(aq) −→ 2H2 O(l)
We find the equivalence point by equating the moles of base with the moles of acid.
moles base = moles acid
(0.1000 L)(0.100 moles OH− /L) = (1.00 moles H+ /L)x
x = 0.01000 L
We see that the equivalence point is reached at 10.0 mL. Since this is a strong/strong
titration, the pH = 7.00 at the equivalence point. That means that 5.0 is pre-equivalence, and
12.0 mL is post-equivalence. In the pre-equivalence region, the added H3 O+ is neutralized
by the OH− in solution. However, not all of the OH− has reacted, so the pH is determined
by how much OH− is left. In the post-equivalence region, all the OH− has reacted, and now
we’re just adding excess H3 O+ ; the pH is determined by how much extra H3 O+ we’ve added.
We go through the calculation below, but first we calculate the total moles of base before
the titration begins.
100.0 mL ×
1 mole OH−
0.100 mole NaOH
×
= 0.01000 mole OH−
1000 mL
1 mole NaOH
At 5.0 mL, we need to calculate how many moles of H3 O+ have been added and then
subtract that from the total moles of hydroxide.
1.00 mole HBr 1 mole H3 O+
×
= 0.00500 mole H3 O+
5.0 mL ×
1000 mL
1 mole HBr
Because of the 1:1 stoichiometry of the titration reaction, the moles of OH− remaining
is 0.01000 − 0.00500 mole = 0.00500 mole. To get [OH− ], we divide by the new total volume
which is 105.0 mL, or 0.1050 L. Thus, [OH− ] = 0.0476 M, pOH = 1.322 , and pH = 12.68.
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At 12.0 mL, we’re in the post-equivalence region, so we need to calculate the excess
H3 O+ that’s been added. Since the equivalence point is at 10.0 mL, we are 2.0 mL past. We
calculate the moles of H3 O+ in 2.0 mL and divide by the total volume.
1.00 mole HBr 1 mole H3 O+
×
= 0.00200 mole H3 O+
2.0 mL ×
1000 mL
1 mole HBr
The new total volume is now 112.0 mL, or 0.1120 L, so [H3 O+ ] = 0.0179 M, and the pH
= 1.75.
R28. Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN
with (a) 4.20 mL of 0.438 M HClO4 , (b) 11.82 of 0.438 M HClO4 , (c) What is
the pH at the equivalence point with 0.438 M HClO4 ?
This is essentially a titration problem, but it’s just been worded differently. First,
we write the equation that describes the reaction between the HClO4 and NaCN. When
these are placed in solution, they dissociate completely. The only active species that we’re
concerned with are the H+ and CN− .
+
CN−
(aq) + H(aq) −→ HCN(aq)
The next step is to find the equivalence point.
moles base = moles acid
(0.05000 L)(0.100 moles CN− /L) = (0.438 moles H+ /L)x
x = 0.01142 L
Now we can identify (a) as being in the pre-equivalence region. In this region, we have
a buffer problem because although some CN− has reacted to form HCN, there’s still some
left. These two species in solution form the components for a buffer. We set up a chart to
organize the numbers.
Reaction:
CN−
(aq)
Initial rel. [ ]:
1
Change: −4.20/11.42
Final rel. [ ]: 1 − 4.20/11.42
+
H+
(aq)
0
4.20/11.42
0
−→
HCN(aq)
0
4.20/11.42
4.20/11.42
We use these relative concentrations in the Henderson-Hasselbalch equation. We get pKa
from the back of the book; it equals 9.208 (Ka = 6.2 × 10−10 ).
[CN− ]
pH = pKa + log
[HCN]
1 − 4.20/11.42
= 9.208 + log
4.20/11.42
= 9.443
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We identify (b) as being in the post-equivalence region. The pH is determined by the
excess strong acid that has been added; we ignore the contribution of HCN to the acidity.
0.4 mL ×
0.438 mole HClO4
1 mole H3 O+
1
×
×
= 0.002834 M H3 O+
1000 mL
1 mole HClO4 0.06182 L
Thus, the pH = 2.5.
For (c), at the equivalence point, we’ve transformed all the CN− to HCN. Now it’s a
weak acid problem. We just need to find the [HCN] and solve it as a weak acid problem.
50.00 mL×
1
0.100 mole NaCN 1 mole CN− 1 mole HCN
×
×
= 0.08141 M HCN
−×
1000 mL
1 mole NaCN 1 mole CN
0.06142 L
This concentration goes in as the initial concentration in our acid equilibrium chart for
HCN.
Reaction: HCN(aq) + H2 O(l)
Initial
0.08141
—
Change
−x
—
Equil. 0.08141 − x
—
*
) H3 O+
(aq)
0
+x
x
+
CN−
(aq)
0
+x
x
[H3 O+ ][CN− ]
[HCN]
x2
6.2 × 10−10 =
0.08143 − x
x = 7.11 × 10−6
Ka =
Using the approximation, [H+ ] = 7.11 × 10−6 (checking the approximation yields 0.009%)
and the pH = 5.15.
R32. Calculate the molar solubility of AgCl in 0.050 M AlCl3 .
This is a common ion effect problem because of the common ion Cl− found in both the
AgCl and the AlCl3 . We need to realize that the 0.050 M AlCl3 dissolves as 0.050 M Al3+
and 0.150 M Cl− . We identify the solubility reaction and its Ksp expression.
−
AgCl(s) *
) Ag+
(aq) + Cl(aq)
Ksp = [Ag+ ][Cl− ]
Now we simply fill in the equilibrium table.
Reaction:
Initial
Change
Equil.
AgCl(s)
–
–
–
*
) Ag+
(aq)
0
+S
S
6
+
Cl−
(aq)
0.150
+S
0.150 + S
We use the equilibrium row to solve for S in the Ksp expression. We look up the Ksp
value for AgCl(s) , and it is 1.8 × 10−10 .
Ksp
1.8 × 10−10
1.8 × 10−10
S
[Ag+ ][Cl− ]
(S)(0.150 + S)
(S)(S)
1.34 × 10−5
=
=
=
=
Given the small value of x, it looks like we’re well within the 5% requirement (it comes
out to 0.009%). Since S and AgCl( s) are in a 1:1 ratio, the molar solubility of AgCl is equal
to x. Therefore, the molar solubility of AgCl(s) is 1.3 × 10−5 M in 0.050 M AlCl3 .
R36. Would a precipitate of silver acetate form if 22.0 mL of 0.100 M AgNO3
were added to 45.0 mL of 0.0260 M NaC2 H3 O2 ? For AgC2 H3 O2 , Ksp = 2.3 ×
10−3 .
This is a Q expression problem; we need to calculate Q and compare its value to Ksp .
The solubility equation we’re interested in is:
−
AgC2 H3 O2(s) *
) Ag+
(aq) + C2 H3 O2(aq)
The Q expression is then:
Q = [Ag+ ][C2 H3 O−
2]
Now we calculate the concentrations of each ion to put into the Q expression.
0.0220 L ×
0.0450 L ×
1
0.100 mole Ag+
×
= 0.03284 M Ag+
1L
0.0670 L
0.0260 mole C2 H3 O−
1
2
×
= 0.01746 M C2 H3 O−
2
1L
0.0670 L
Now we put these concentrations into the Q expression.
Q = [Ag+ ][C2 H3 O−
2]
= (0.03284 )(0.01746 )
= 5.734 × 10−4
We see that 5.734 × 10−4 < 2.3 × 10−3 , which means that Q< Ksp . Therefore, a
precipitate will not form.
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