Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Can you help me calculate the concentrations for 4&5 and explain to me what effect did NH4+ have on the equilibrium/ph. Also, can you explain to me why the pH change a smaller for solutions 4&5 then it is in solutions 1&2.
![SOL
ER
Equilibria with Weak Acids & Weak Bases - Results
Exp
Composition
1
A few mL of 0.10M NH3 (unchanged)
2
2.5mL 0.1M NH3 & 7.5mL H₂O
3
2.5mL 0.1M NH4CI & 7.5mL H₂O
4
2.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mL
H₂O
Take 1mL Solution #4 and add 9mL of water
GEODO
Iziprasidons
5
10 ml
Solution 1. Use the concentration of NH3 and the measured pH above to
determine the Kb of NH3. (show the ICE table and calculations)
-14
PH= PKa + log ([A]/[B]
[AH□ CH]
9.26- Pla+log (NHG]
(NH4)
9.24 = Pleat log (a']
(0.0
Pla = 9,26
pkg==logka
Ka=-antilog Pla
ках кв = 10
кв = 10-14
Ka
10-14
1.81x109
Total Vol
n/a
10 ml
10 ml
10 ml
1.81
M₂ = (2.5x0.1)/10
M2 = 0.025M
K6=5,55 x10-24
Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ?
-Antilog (9.26) = -1.81×1091
X 10
-23
1
pH
11.13
10.82
6.01
9.26
9.26
Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3.
M₁ V₁ = M₂V2
Conc. of NH3
pH measured: 10.82
2.5 X0,1 = M2 X10mL
m² = 2.5x0.¹ = 0,023M
Compare solutions 1 & 2. What effect did diluting the base have on the pH?
The addition of H₂0, slightly decreased the pH.
Use the new concentration and pH to measure the Kb of NH3 again. Did it
change?
Kb will not change til the temp changes
Kb=cok²
x²=1,8225x105
0.025
α = 0,027
Solution 3 Conc. of NH4+
pH of solution 3: 6.01
Use the diluted concentration of NH4+ and measured pH to determine the Ka of
NH4+
M₁ V₁ = m₂ V₂
61 = 1/2 p²
pk
2.5ml (0.1M) = M₂ (10m²)
[OH-]=CQ=0,025x0,021
-log[04²] = -log[0,025 x =
0,027)
POH= 3,17
PH=14-3₁17=1085
17
PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10"""
6.01 = pka +10g ( [0.137 (0.17)
Kb = 1x10" / Ka
pka = 6,01 X10"
Ka=-1.02 x 109
Kb = 1x10"¹"/1.02X10²
Kb = (1/1.02)×10 ²3
-23
Kb= 0.9804 X16-23
K5= 9.804 X10-24
9
How does th
good) This
Calculate th
Sol
2
4
5
Compare
equilibriu
Com](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F154975ff-ecfa-4e64-bed9-4b9b79aed696%2F9256be6c-cd58-4450-b42d-940ba31787f7%2Fpsutawx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:SOL
ER
Equilibria with Weak Acids & Weak Bases - Results
Exp
Composition
1
A few mL of 0.10M NH3 (unchanged)
2
2.5mL 0.1M NH3 & 7.5mL H₂O
3
2.5mL 0.1M NH4CI & 7.5mL H₂O
4
2.5mL 0.1M NH3 & 2.5mL 0.1M NH4CI & 5mL
H₂O
Take 1mL Solution #4 and add 9mL of water
GEODO
Iziprasidons
5
10 ml
Solution 1. Use the concentration of NH3 and the measured pH above to
determine the Kb of NH3. (show the ICE table and calculations)
-14
PH= PKa + log ([A]/[B]
[AH□ CH]
9.26- Pla+log (NHG]
(NH4)
9.24 = Pleat log (a']
(0.0
Pla = 9,26
pkg==logka
Ka=-antilog Pla
ках кв = 10
кв = 10-14
Ka
10-14
1.81x109
Total Vol
n/a
10 ml
10 ml
10 ml
1.81
M₂ = (2.5x0.1)/10
M2 = 0.025M
K6=5,55 x10-24
Use the equation (Ka*Kb = 1*10-14) to calculate the Ka for NH4* ?
-Antilog (9.26) = -1.81×1091
X 10
-23
1
pH
11.13
10.82
6.01
9.26
9.26
Solution 2 Use the equation M₁V₁= M₂V₂ to calculate new conc. of NH3.
M₁ V₁ = M₂V2
Conc. of NH3
pH measured: 10.82
2.5 X0,1 = M2 X10mL
m² = 2.5x0.¹ = 0,023M
Compare solutions 1 & 2. What effect did diluting the base have on the pH?
The addition of H₂0, slightly decreased the pH.
Use the new concentration and pH to measure the Kb of NH3 again. Did it
change?
Kb will not change til the temp changes
Kb=cok²
x²=1,8225x105
0.025
α = 0,027
Solution 3 Conc. of NH4+
pH of solution 3: 6.01
Use the diluted concentration of NH4+ and measured pH to determine the Ka of
NH4+
M₁ V₁ = m₂ V₂
61 = 1/2 p²
pk
2.5ml (0.1M) = M₂ (10m²)
[OH-]=CQ=0,025x0,021
-log[04²] = -log[0,025 x =
0,027)
POH= 3,17
PH=14-3₁17=1085
17
PH=pKa + log ((A] [B])/([AH] [BH]) ( KaxKb = 1x10"""
6.01 = pka +10g ( [0.137 (0.17)
Kb = 1x10" / Ka
pka = 6,01 X10"
Ka=-1.02 x 109
Kb = 1x10"¹"/1.02X10²
Kb = (1/1.02)×10 ²3
-23
Kb= 0.9804 X16-23
K5= 9.804 X10-24
9
How does th
good) This
Calculate th
Sol
2
4
5
Compare
equilibriu
Com
Transcribed Image Text:70°
27
b/c we are
antiloy to
How does this Ka value compare to the value on pg. 1? (within a power of 10 is
Calculate ting value what the
good) This
Calculate the concentrations for solution 4 and solution 5 value is.
Sol
Conc. NH3 (M)
Conc.NH4+ (M)
0.025M
2
0.025M
pH
10.82
9.26
9.26
4
5
Compare pH of solutions 2&4. What effect did adding NH4* have on the
equilibrium/pH?
Compare the pH from dilution in solutions 1&2 to the pH change in solutions 4 &
5. Why is the pH change smaller for solutions 4&5 than in solutions 1&2? Think
above the ratio of base and conjugate acid. You can also show this
mathematically by using the Kb from part 1 with NH3/NH4* conjugates to
determine what the OH- concentration should be.
2
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