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CHAPTER 15
ACID-BASE EQUILIBRIA
Questions
- When an acid dissociates, ions are produced. The common ion effect is observed when one of the product ions in a particular equilibrium is added from an outside source. For a weak acid dissociating to its conjugate base and H+, the common ion would be the conjugate base; this would be added by dissolving a soluble salt of the conjugate base into the acid solution. The presence of the conjugate base from an outside source shifts the equilibrium to the left so less acid dissociates.
- pH = pKa + log [acid]
[base]; when [acid] > [base], then [acid]
[base]< 1 and log
[acid ]
[base] < 0.
From the Henderson-Hasselbalch equation, if the log term is negative, then pH < pKa. When one has more acid than base in a buffer, the pH will be on the acidic side of the pKa value; that is, the pH is at a value lower than the pKa value. When one has more base than acid in a buffer ([conjugate base] > [weak acid]), then the log term in the Henderson-Hasselbalch equation is positive, resulting in pH > pKa. When one has more base than acid in a buffer, the pH is on the basic side of the pKa value; that is, the pH is at a value greater than the pKa value. The other scenario you can run across in a buffer is when [acid] = [base]. Here, the log term is equal to zero, and pH = pKa.
- The more weak acid and conjugate base present, the more H+^ and/or OH^ that can be absorbed by the buffer without significant pH change. When the concentrations of weak acid and conjugate base are equal (so that pH = pKa), the buffer system is equally efficient at absorbing either H+^ or OH. If the buffer is overloaded with weak acid or with conjugate base, then the buffer is not equally efficient at absorbing either H+^ or OH.
- a. The red plot is the pH curve for the strong acid and the blue plot is the pH curve for the weak acid. The pH at the equivalence point is 7.00 for the strong acid-strong base titration, while the pH is greater than 7.00 if a weak acid is titrated. Another point one could look at is the initial point. Because both acids have the same concentration, the strong acid curve will be at the lowest initial pH. Actually, any point at any volume up to the equivalence point for the strong acid plot will have a lower pH than the weak acid plot (assuming equal concentrations and volumes). Another difference would be the pH at the halfway point to equivalence. For the weak acid titration, the pH of solution equals the pKa value for the weak acid at the halfway point to equivalence; this is not the case when a strong acid is titrated. b. A buffer is a solution that resists pH change. From this definition, both titrations have regions where the pH doesn’t change much on addition of strong base, so both could be labeled to have buffer regions. However, we don’t normally include strong acids as a component of buffer solutions. Strong acids certainly can absorb added OH^ by reacting
with it to form water. But when more strong acid is added, the H+^ concentration increases steadily; there is nothing present in a strong acid solution to react with added H+.
This is not the case in the weak acid-strong base titration. After some OH^ has been added, some weak acid is converted into its conjugate base. We now have a typical buffer solution because there are significant amounts of weak acid and conjugate base present at the same time. The buffer region extends from a little past the initial point in the titration up to just a little before the equivalence point. This entire region is a buffer region because both the weak acid and conjugate base are present in significant quantities in this region. c. True; HA + OH^ A^ + H 2 O; both reactions have the same neutralization reaction. In both cases, the equivalence point is reached when enough OH^ has been added to exactly react with the acid present initially. Because all acid concentrations and volumes are the same, we have equal moles of each acid which requires the same moles of OH^ to reach the equivalence point. Therefore, each acid requires the same volume of 0.10 M NaOH to reach the equivalence point. d. False; the pH for the strong acid-strong base titration will be 7.00 at the equivalence point. The pH for the weak acid-strong base titration will be greater than 7.00 at the equivalence point. In both titrations, the major species present at the equivalence points are Na+, H 2 O, and the conjugate base of the acid titrated. Because the conjugate base of a strong acid has no basic characteristics, pH = 7.00 at the equivalence point. However, the conjugate base of a weak acid is a weak base. A weak base is present at the equivalence point of a weak acid-strong base titration, so the pH is basic (pH > 7.0).
- a. Let’s call the acid HA, which is a weak acid. When HA is present in the beakers, it exists in the undissociated form, making it a weak acid. A strong acid would exist as separate H+^ and A^ ions.
b. Beaker a contains 4 HA molecules and 2 A^ ions, beaker b contains 6 A^ ions, beaker c contains 6 HA molecules, beaker d contains 6 A^ and 6 OH^ ions, and beaker e contains 3 HA molecules and 3 A^ ions. HA + OH^ A^ + H 2 O; this is the neutralization reaction that occurs when OH^ is added. We start off the titration with a beaker full of weak acid (beaker c). When some OH^ is added, we convert some weak acid HA into its conjugate base A^ (beaker a). At the halfway point to equivalence, we have converted exactly one-half of the initial amount of acid present into its conjugate base (beaker e). We finally reach the equivalence point when we have added just enough OH^ to convert all of the acid present initially into its conjugate base (beaker b). Past the equivalence point, we have added an excess of OH, so we have excess OH^ present as well as the conjugate base of the acid produced from the neutralization reaction (beaker d). The order of the beakers from start to finish is: beaker c beaker a beaker e beaker b beaker d
c. pH = pKa when a buffer solution is present that has equal concentrations of the weak acid and conjugate base. This is beaker e. d. The equivalence point is when just enough OH^ has been added to exactly react with all of the acid present initially. This is beaker b.
From the pKa values, the correct ordering at the halfway point to equivalence would be i < iii < iv < ii. Note that for the weak base-strong acid titration using C 5 H 5 N, the pH is acidic at the halfway point to equivalence, whereas the weak acid-strong base titration using HOC 6 H 5 is basic at the halfway point to equivalence. This is fine; this will always happen when the weak base titrated has a Kb < 1 × 10 ^7 (so Ka of the conjugate acid is greater than 1 × 10^7 ) and when the weak acid titrated has a Ka < 1 × 10 ^7 (so Kb of the conjugate base is greater than 1 × 10^7 ).
- The three key points to emphasize in your sketch are the initial pH, the pH at the halfway point to equivalence, and the pH at the equivalence point. For all the weak bases titrated, pH = pKa at the halfway point to equivalence (50.0 mL HCl added) because [weak base] = [conjugate acid] at this point. Here, the weak base with Kb = 1 10 ^5 has a conjugate acid with Ka = 1 10 ^9 , so pH = 9.0 at the halfway point. The weak base with Kb = 1 10 ^10 has a pH = 4.0 at the halfway point to equivalence. For the initial pH, the strong base has the highest pH (most basic), whereas the weakest base has the lowest pH (least basic). At the equivalence point (100.0 mL HCl added), the strong base titration has pH = 7.0. The weak bases titrated have acidic pH’s because the conjugate acids of the weak bases titrated are the major species present. The weakest base has the strongest conjugate acid so its pH will be lowest (most acidic) at the equivalence point.
Volume HCl added (mL)
- HIn ⇌ H+^ + In^ Ka = [HIn]
[H ][In]
Indicators are weak acids themselves. The special property they have is that the acid form of the indicator (HIn) has one distinct color, whereas the conjugate base form (In) has a different distinct color. Which form dominates and thus determines the color of the solution is determined by the pH. An indicator is chosen in order to match the pH of the color change at about the pH of the equivalence point.
Kb = 10-
Kb = 10-
strong base
pH 7.
Exercises
Buffers
- Only the third (lower) beaker represents a buffer solution. A weak acid and its conjugate base must both be present in large quantities in order to have a buffer solution. This is only the case in the third beaker. The first beaker represents a beaker full of strong acid which is 100% dissociated. The second beaker represents a weak acid solution. In a weak acid solution, only a small fraction of the acid is dissociated. In this representation, 1/10 of the weak acid has dissociated. The only B^ present in this beaker is from the dissociation of the weak acid. A buffer solution has B^ added from another source.
- A buffer solution is a solution containing a weak acid plus its conjugate base or a weak base plus its conjugate acid. Solution c contains a weak acid (HOCl) plus its conjugate base (OCl), so it is a buffer. Solution e is also a buffer solution. It contains a weak base (H 2 NNH 2 ) plus its conjugate acid (H 2 NNH 3 +).
Solution a contains a strong acid (HBr) and a weak acid (HOBr). Solution b contains a strong acid (HClO 4 ) and a strong base (RbOH). Solution d contains a strong base (KOH) and a weak base (HONH 2 ).
- When strong acid or strong base is added to a bicarbonate-carbonate mixture, the strong acid(base) is neutralized. The reaction goes to completion, resulting in the strong acid(base) being replaced with a weak acid(base), resulting in a new buffer solution. The reactions are:
H+(aq) + CO 32 (aq) HCO 3 (aq); OH^ + HCO 3 (aq) CO 32 (aq) + H 2 O(l)
- Similar to the HCO 3 /CO 32 ^ buffer discussed in Exercise 19, the HONH 3 +/HONH 2 buffer absorbs added OH^ and H+^ in the same fashion.
HONH 2 (aq) + H+(aq) HONH 3 +(aq)
HONH 3 +(aq) + OH(aq) HONH 2 (aq) + H 2 O(l)
- a. This is a weak acid problem. Let HC 3 H 5 O 2 = HOPr and C 3 H 5 O 2 ^ = OPr.
HOPr(aq) ⇌ H+(aq) + OPr(aq) Ka = 1.3 × 10^5 Initial 0.100 M ~0 0 x mol/L HOPr dissociates to reach equilibrium Change x + x + x Equil. 0.100 x x x
Ka = 1.3 × 10^5 = [HOPr] 0. 100 0. 100
[H ][OPr]^2 x^2 x
x
x = [H+] = 1.1 × 10^3 M ; pH = 2.96; assumptions good by the 5% rule.
Kb = 1.1 × 10 ^8 = x
x
- 100
2
- 100
x^2
x = [OH] = 3.3 × 10 ^5 M ; pOH = 4.48; pH = 9.52; assumptions good.
b. Weak acid problem (Cl^ has no acidic/basic properties);
HONH 3 +^ ⇌ HONH 2 + H+ Initial 0.100 M 0 ~ x mol/L HONH 3 +^ dissociates to reach equilibrium
Change – x + x + x
Equil. 0.100 – x x x
Ka = b
w K
K
= 9.1 × 10 ^7 =
[HONH ]
[HONH ][H ]
3
2
x
x
- 100
2
- 100
x^2
x = [H+] = 3.0 × 10 ^4 M ; pH = 3.52; assumptions good.
c. Pure H 2 O, pH = 7. d. Buffer solution where pKa = – log(9.1 × 10 7 )= 6.04. Using the Henderson-Hasselbalch equation:
pH = pKa + log [acid]
[base] = 6.04 + log [HONH ]
[HONH]
3
2 (^) = 6.04 + log( 0. 100 )
- 0.100 M HC 3 H 5 O 2 : Percent dissociation = [HC 3 H 5 O 2 ] 0
[H ] × 100 = M
M
100
1 10 ^3 × 100
= 1.1%
0.100 M HC 3 H 5 O 2 + 0.100 M NaC 3 H 5 O 2 : % dissociation =
- 100
1. 3 10 ^5
× 100 = 1.3 × 10 ^2 %
The percent dissociation of the acid decreases from 1.1% to 1.3 × 10 ^2 % (a factor of 85) when C 3 H 5 O 2 ^ is present. This is known as the common ion effect. The presence of the conjugate base of the weak acid inhibits the acid dissociation reaction.
- 0.100 M HONH 2 : Percent ionization [HONH 2 ] 0
[OH ] × 100 = M
M
100
3 10 ^5 × 100
= 3.3 × 10 ^2 %
0.100 M HONH 2 + 0.100 M HONH 3 +: % ionization =
- 100
1. 1 10 ^8
× 100 = 1.1 × 10 ^5 %
The percent ionization decreases by a factor of 3000. The presence of the conjugate acid of the weak base inhibits the weak base reaction with water. This is known as the common ion effect.
- a. We have a weak acid (HOPr = HC 3 H 5 O 2 ) and a strong acid (HCl) present. The amount of H+^ donated by the weak acid will be negligible. To prove it, consider the weak acid equilibrium reaction:
HOPr ⇌ H+^ + OPr^ Ka = 1.3 × 10^5 Initial 0.100 M 0.020 M 0 x mol/L HOPr dissociates to reach equilibrium Change x + x + x Equil. 0.100 x 0.020 + x x
[H+] = 0.020 + x 0.020 M ; pH = 1.70; assumption good ( x = 6.5 × 10^5 is << 0.020).
Note : The H+^ contribution from the weak acid HOPr was negligible. The pH of the solution can be determined by only considering the amount of strong acid present.
b. Added H+^ reacts completely with the best base present, OPr.
OPr^ + H+^ HOPr Before 0.100 M 0.020 M 0 Change 0.020 0.020 +0.020 Reacts completely After 0.080 0 0.020 M After reaction, a weak acid, HOPr , and its conjugate base, OPr, are present. This is a buffer solution. Using the Henderson-Hasselbalch equation where pKa = log (1.3 × 10 ^5 ) = 4.89:
pH = pKa + log [acid]
[base] = 4.89 + log ( 0. 020 )
= 5.49; assumptions good.
c. This is a strong acid problem. [H+] = 0.020 M ; pH = 1.
d. Added H+^ reacts completely with the best base present, OPr.
OPr^ + H+^ HOPr Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020 +0.020 Reacts completely After 0.080 0 0.
A buffer solution results (weak acid + conjugate base). Using the Henderson- Hasselbalch equation:
pH = pKa + log [acid]
[base] = 4.89 + log ( 0. 120 )
= 4.71; assumptions good.
- a. Added H+^ reacts completely with HONH 2 (the best base present) to form HONH 3 +.
HONH 2 + H+^ HONH 3 +
Before 0.100 M 0.020 M 0
Change – 0.020 – 0.020 +0.020 Reacts completely
After 0.080 0 0.
[OH] = 0.020 + x 0.020 M ; pOH = 1.70; pH = 12.30; assumption good.
Note : The OH^ contribution from the weak base OPr^ was negligible ( x = 3.9 × 10^9 M as compared to 0.020 M OH-^ from the strong base). The pH can be determined by only con- sidering the amount of strong base present.
c. This is a strong base in water. [OH] = 0.020 M ; pOH = 1.70; pH = 12.
d. OH^ will react completely with HOPr, the best acid present.
HOPr + OH^ OPr^ + H 2 O Before 0.100 M 0.020 M 0.100 M Change 0.020 0.020 +0.020 Reacts completely After 0.080 0 0.
Using the Henderson-Hasselbalch equation to solve for the pH of the resulting buffer solution:
pH = pKa + log [acid]
[base] = 4.89 + log ( 0. 080 )
= 5.07; assumptions good.
- a. We have a weak base and a strong base present at the same time. The OH^ contribution from the weak base, HONH 2 , will be negligible. Consider only the added strong base as the primary source of OH.
[OH] = 0.020 M ; pOH = 1.70; pH = 12.
b. Added strong base will react to completion with the best acid present, HONH 3 +.
OH^ + HONH 3 +^ HONH 2 + H 2 O
Before 0.020 M 0.100 M 0
Change – 0.020 – 0.020 +0.020 Reacts completely
After 0 0.080 0.
The resulting solution is a buffer (a weak acid and its conjugate base). Using the Henderson-Hasselbalch equation:
pH = 6.04 + log ( 0. 080 )
c. This is a strong base in water. [OH] = 0.020 M ; pOH = 1.70; pH = 12.
d. Major species: H 2 O, Cl, Na+, HONH 2 , HONH 3 +, OH; again, the added strong base reacts completely with the best acid present, HONH 3 +.
HONH 3 +^ + OH^ HONH 2 + H 2 O Before 0.100 M 0.020 M 0.100 M Change – 0.020 – 0.020 +0.020 Reacts completely After 0.080 0 0.
A buffer solution results. Using the Henderson-Hasselbalch equation:
pH = 6.04 + log [HONH ]
[HONH ]
3
2 (^) = 6.04 + log( 0. 080 )
- Consider all the results to Exercises 21, 25, and 27:
Solution Initial pH After Added H+^ After Added OH a 2.96 1.70 4. b 8.94 5.49 12. c 7.00 1.70 12. d 4.89 4.71 5.
The solution in Exercise 21d is a buffer; it contains both a weak acid (HC 3 H 5 O 2 ) and a weak base (C 3 H 5 O 2 ). Solution d shows the greatest resistance to changes in pH when either a strong acid or a strong base is added, which is the primary property of buffers.
- Consider all of the results to Exercises 22, 26, and 28.
Solution Initial pH After Added H+^ After Added OH a 9.52 6.64 12. b 3.52 1.70 5. c 7.00 1.70 12. d 6.04 5.86 6.
The solution in Exercise 22d is a buffer; it shows the greatest resistance to a change in pH when strong acid or base is added. The solution in Exercise 22d contains a weak acid (HONH 3 +) and a weak base (HONH 2 ), which constitutes a buffer solution.
- Major species: HNO 2 , NO 2 ^ and Na+. Na+^ has no acidic or basic properties. One appropriate equilibrium reaction you can use is the Ka reaction of HNO 2 , which contains both HNO 2 and NO 2 . However, you could also use the Kb reaction for NO 2 ^ and come up with the same answer. Solving the equilibrium problem (called a buffer problem):
HNO 2 ⇌ NO 2 ^ + H+ Initial 1.00 M 1.00 M ~ x mol/L HNO 2 dissociates to reach equilibrium Change – x + x + x Equil. 1.00 – x 1.00 + x x
Ka = 4.0 × 10 ^4 = [HNO ]
[NO ][H ]
2
2
x
x x
- 00
( 1. 00 )( x ) (assuming x << 1.00)
x = 4.0 × 10 ^4 M = [H+]; assumptions good ( x is 4.0 × 10 ^2 % of 1.00).
pH = – log(4.0 × 10 ^4 )= 3.
Note : We would get the same answer using the Henderson-Hasselbalch equation. Use whichever method you prefer.
Major species after HCl added: HNO 2 , NO 2 , H+, Na+, Cl; the added H+^ from the strong acid will react completely with the best base present (NO 2 ).
H+^ + NO 2 ^ HNO 2
Before
00 L
20 mol 1.00 M 1.00 M
Change – 0.20 M – 0.20 M +0.20 M Reacts completely
After 0 0.80 1.
After all the H+^ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem:
HNO 2 ⇌ NO 2 ^ + H+ Initial 1.20 M 0.80 M 0 Equil. 1.20 – x 0.80 + x + x
Ka = 4.0 × 10 ^4 = x
x x
( 0. 80 )( x ) , x = [H+] = 6.0 × 10 ^4 M ; pH = 3.22;
assumptions good.
Note : The added HCl to this buffer solution changes the pH only from 3.40 to 3.22. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70.
- Major species after NaOH added: HF, F, K+, Na+, OH, and H 2 O. The OH^ from the strong base will react with the best acid present (HF). Any reaction involving a strong base is assumed to go to completion. Because all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is:
OH^ + HF F^ + H 2 O
Before 0.10 mol/1.00 L 0.60 M 1.00 M
Change – 0.10 M – 0.10 M +0.10 M Reacts completely
After 0 0.50 1.
After all the OH^ reacts, we are left with a solution containing a weak acid (HF) and its conjugate base (F). This is what we call a buffer problem. We will solve this buffer problem using the Ka equilibrium reaction. One could also use the Kb equilibrium reaction or use the Henderson-Hasselbalch equation to solve for the pH.
HF ⇌ F^ + H+
Initial 0.50 M 1.10 M ~ x mol/L HF dissociates to reach equilibrium
Change – x + x + x
Equil. 0.50 – x 1.10 + x x
Ka = 7.2 × 10 ^4 = x
x x
( 1. 10 )( x ) , x = [H+] = 3.3 × 10 ^4 M ; pH = 3.48;
assumptions good.
Note : The added NaOH to this buffer solution changes the pH only from 3.37 to 3.48. If the NaOH were added to 1.0 L of pure water, the pH would change from 7.00 to 13.00. Major species after HCl added: HF, F, H+, K+, Cl, and H 2 O; the added H+^ from the strong acid will react completely with the best base present (F).
H+^ + F^ HF
Before
00 L
20 mol 1.00 M 0.60 M
Change – 0.20 M – 0.20 M +0.20 M Reacts completely
After 0 0.80 0. After all the H+^ has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem:
HF ⇌ F^ + H+
Initial 0.80 M 0.80 M 0 Equil. 0.80 – x 0.80 + x x
Ka = 7.2 × 10 ^4 = x
x x
( 0. 80 )( x ) , x = [H+] = 7.2 × 10 ^4 M ; pH = 3.14;
assumptions good.
Note : The added HCl to this buffer solution changes the pH only from 3.37 to 3.14. If the HCl were added to 1.0 L of pure water, the pH would change from 7.00 to 0.70.
35. a. HC 2 H 3 O 2 ⇌ H+^ + C 2 H 3 O 2 −^ Ka = 1.8 × 10 ^5
Initial 0.10 M ~0 0.25 M x mol/L HC 2 H 3 O 2 dissociates to reach equilibrium Change x + x + x Equil. 0.10 x x 0.25 + x
1.8 × 10^5 =
( 0. 25 x x)
x x)
(assuming 0.25 + x 0.25 and 0.10 – x 0.10)
x = [H+] = 7.2 × 10^6 M ; pH = 5.14; assumptions good by the 5% rule.
Alternatively, we can use the Henderson-Hasselbalch equation:
pH = pKa + log , [acid]
[base] where pKa = log(1.8 × 10^5 ) = 4.
pH = 4.74 + log ( 0. 10 )
The Henderson-Hasselbalch equation will be valid when assumptions of the type, 0.10 x 0.10, that we just made are valid. From a practical standpoint, this will almost always be true for useful buffer solutions. Note : The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions.
Alternatively, we can use the Henderson-Hasselbalch equation to calculate the pH of buffer solutions.
pH = pKa + log [acid]
[base] = pKa + log [HCHO]
[CHO ]
7 5 2
7 5 2
pH = log(6.4 × 10 ^5 ) + log
Within round-off error, this is the same answer we calculated solving the equilibrium problem using the Ka reaction.
The Henderson-Hasselbalch equation will be valid when an assumption of the type 1.31 + x 1.31 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the pH. Note : The Henderson-Hasselbalch equation can only be used to solve for the pH of buffer solutions.
- 50.0 g NH 4 Cl ×
- 49 gNHCl
1 molNHCl 4
4 = 0.935 mol NH 4 Cl added to 1.00 L; [NH 4 +] = 0.935 M
Using the Henderson Hasselbalch equation to solve for the pH of this buffer solution:
pH = pKa + log [NH ]
[NH] 4
3 (^) = log(5.6 × 10 ^10 ) + log
[H+] added =
- 2500 L
010 mol = 0.040 M ; the added H+^ reacts completely with NH 3 to form NH 4 +.
a. NH 3 + H+^ NH 4 + Before 0.050 M 0.040 M 0.15 M Change 0.040 0.040 +0.040 Reacts completely After 0.010 0 0.
A buffer solution still exists after H+^ reacts completely. Using the Henderson- Hasselbalch equation:
pH = pKa + log [NH ]
[NH] 4
3 (^) = log(5.6 × 10 (^10) ) + log
b. NH 3 + H+^ NH 4 + Before 0.50 M 0.040 M 1.50 M Change 0.040 0.040 +0.040 Reacts completely After 0.46 0 1.
A buffer solution still exists. pH = pKa + log [NH ]
[NH]
4
3 (^) , 9.25 + log
The two buffers differ in their capacity and not their initial pH (both buffers had an initial pH = 8.77). Solution b has the greatest capacity since it has the largest concentrations of weak acid and conjugate base. Buffers with greater capacities will be able to absorb more added H+^ or OH.
- a. pKb for C 6 H 5 NH 2 = log(3.8 × 10^10 ) = 9.42; pKa for C 6 H 5 NH 3 +^ = 14.00 9.42 = 4.
pH = pKa + log [CHNH ]
[CHNH ]
6 5 3
6 5 2 (^) , 4.20 = 4.58 + log[CHNH ]
6 5 3
M
0.38 = log [CHNH ]
6 5 3
M
, [C 6 H 5 NH 3 +] = [C 6 H 5 NH 3 Cl] = 1.2 M
b. 4.0 g NaOH × molNaOH
1 molOH
- 00 g
1 molNaOH = 0.10 mol OH; [OH] =
0 L
10 mol = 0.10 M
C 6 H 5 NH 3 +^ + OH^ C 6 H 5 NH 2 + H 2 O
Before 1.2 M 0.10 M 0.50 M Change 0.10 0.10 +0.10 Reacts completely After 1.1 0 0.
A buffer solution exists. pH = 4.58 + log
- pH = pKa + log [HCHO ]
[CHO ]
2 3 2
2 3 2
; pKa = – log(1.8 × 10 ^5 )= 4.
Because the buffer components, C 2 H 3 O 2 ^ and HC 2 H 3 O 2 , are both in the same volume of solution, the concentration ratio of [C 2 H 3 O 2 - ] : [HC 2 H 3 O 2 ] will equal the mole ratio of mol C 2 H 3 O 2 ^ to mol HC 2 H 3 O 2.
5.00 = 4.74 + log ; molHCHO
molCHO 2 3 2
2 3 2
mol HC 2 H 3 O 2 = 0.5000 L × L
- 200 mol = 0.100 mol
0.26 = log
- 100
molCHO ,
- 100 mol
mol C 2 H 3 O 2 2 3 2 = 100.26^ = 1.8, mol C 2 H 3 O 2 ^ = 0.18 mol
Mass NaC 2 H 3 O 2 = 0.18 mol NaC 2 H 3 O 2 × mol
03 g = 15 g NaC 2 H 3 O 2
pH = pKa + log [HNO ]
[NO ]
2
2
, 3.55 = log(4.0 × 10^4 ) + log [HNO ]
[NO ] 2
2
3.55 = 3.40 + log [HNO]
[NO ]
2
2
, [HNO ]
[NO ] 2
2
= 100.15^ = 1.4 = 2
2 mol HNO
mol NO
c. 10.00 = 9.25 + log [NH ]
[NH]
4
3 (^) , [NH ]
[NH ]
4
3 (^) = 10
d. 9.60 = 9.25 + log [NH ]
[NH]
4
3 (^) , [NH ]
[NH]
4
3 (^) = 10
- pH = pKa + [HCO]
[HCO ]
log 2 3
3
, 7.40 = log(4.3 × 10^7 ) +
- 0012
[HCO ]
log 3
[HCO ]
log 3
= 7.40 – 6.37 = 1.03,
- 0012
[HCO 3 ]
= 101.03, [HCO 3 ] = 1.3 × 10^2 M
- At pH = 7.40: 7.40 = log(4.3 × 10^7 ) + [HCO]
[HCO ]
log 2 3
3
[HCO ]
[HCO ]
log 2 3
3
= 7.40 – 6.37 = 1.03, [HCO ]
[HCO ]
2 3
3
= 101.03, [HCO ]
[HCO ]
3
2 3 (^) = 10
At pH = 7.35: [HCO ]
[HCO ]
log 2 3
3
= 7.35 – 6.37 = 0.98, [HCO ]
[HCO ]
2 3
3
= 100.
[HCO ]
[HCO ]
3
2 3 ^ = 10
0.98^ = 0.
The [H 2 CO 3 ] : [HCO 3 ] concentration ratio must increase from 0.093 to 0.10 in order for the onset of acidosis to occur.
- A best buffer has large and equal quantities of weak acid and conjugate base. Because [acid]
= [base] for a best buffer, pH = pKa + log [acid]
[base] = pKa + 0 = pKa (pH pKa for a best buffer).
The best acid choice for a pH = 7.00 buffer would be the weak acid with a pKa close to 7.0 or Ka 1 × 10^7. HOCl is the best choice in Table 14.2 (Ka = 3.5 × 10^8 ; pKa = 7.46). To make this buffer, we need to calculate the [base] : [acid] ratio.
7.00 = 7.46 + log [HOCl]
[OCl ] , [acid]
[base ] = 100.46^ = 0.
Any OCl/HOCl buffer in a concentration ratio of 0.35 : 1 will have a pH = 7.00. One possibility is [NaOCl] = 0.35 M and [HOCl] = 1.0 M.
- For a pH = 5.00 buffer, we want an acid with a pKa close to 5.00. For a conjugate acid-base pair, 14.00 = pKa + pKb. So, for a pH = 5.00 buffer, we want the base to have a pKb close to (14.0 5.0 =) 9.0 or a Kb close to 1 × 10^9. The best choice in Table 14.3 is pyridine (C 5 H 5 N) with Kb = 1.7 × 10^9.
pH = pKa + log 9
14
b
w a (^1). 7 10
K
K
; K
[acid]
[base ]
= 5.9 × 10^6
5.00 = log(5.9 × 10-6) + log [CHNH ]
[CHN]
[acid]
[base] 5 5
5 5 (^) = 10
0.23^ = 0.
There are many possibilities to make this buffer. One possibility is a solution of [C 5 H 5 N] = 0.59 M and [C 5 H 5 NHCl] = 1.0 M. The pH of this solution will be 5.00 because the base to acid concentration ratio is 0.59 : 1.
- Ka for H 2 NNH 3 +^ = Kw /Kb,H 2 NNH 2 = 1.0 × 10^14 /3.0 × 10^6 = 3.3 × 10^9
pH = pKa + [H NNH ]
[HNNH ]
log 2 3
2 2 (^) = log(3.3 × 10
^9 ) +
log = 8.48 + (0.30) = 8.
pH = pKa for a buffer when [acid] = [base]. Here, the acid (H 2 NNH 3 +) concentration needs to decrease, while the base (H 2 NNH 2 ) concentration needs to increase in order for [H 2 NNH 3 +] = [H 2 NNH 2 ]. Both of these changes are accomplished by adding a strong base (like NaOH) to the original buffer. The added OH^ from the strong base converts the acid component of the buffer into the conjugate base. Here, the reaction is H 2 NNH 3 +^ + OH^ H 2 NNH 2 + H 2 O. Because a strong base is reacting, the reaction is assumed to go to completion. The following set-up determines the number of moles of OH( x ) that must be added so that mol H 2 NNH 3 +^ = mol H 2 NNH 2. When mol acid = mol base in a buffer, then [acid] = [base] and pH = pKa.
H 2 NNH 3 +^ + OH^ H 2 NNH 2 + H 2 O Before 1.0 L × 0.80 mol/L x 1.0 L × 0.40 mol/L Change x x + x Reacts completely After 0.80 x 0 0.40 + x
We want mol H 2 NNH 3 +^ = mol H 2 NNH 2. Therefore:
0.80 x = 0.40 + x , 2 x = 0.40, x = 0.20 mol OH
When 0.20 mol OH^ is added to the initial buffer, mol H 2 NNH 3 +^ is decreased to 0.60 mol, while mol H 2 NNH 2 is increased to 0.60 mol. Therefore, 0.20 mol of NaOH must be added to the initial buffer solution in order to produce a solution where pH = pKa.
- pH = pKa + [HOCl]
[OCl ] log
= log(3.5 × 10^8 ) +
log = 7.46 + 0.65 = 8.
pH = pKa when [HOCl] = [OCl] (or when mol HOCl = mol OCl). Here, the moles of the base component of the buffer must decrease, while the moles of the acid component of the buffer must increase in order to achieve a solution where pH = pKa. Both of these changes occur when a strong acid (like HCl) is added. Let x = mol H+^ added from the strong acid HCl.
