Solution Measured pH Theoretical [H30*| |Calculated pH Distilled water 7.10 Original buffer solution 1. 35 x 105 M 4.87 4.4 Distilled water with added 6M HCI 1.74 0.01819M Buffer solution with 4.41 6.7x 10 5 M added 6M HCI Distilled water with added 6M NaOH 12.26 5.50 x 10 M Buffer solution with 5.12 7.586 15°M added 6M NaOH ow to calculate [H3O*] and pH?
Solution Measured pH Theoretical [H30*| |Calculated pH Distilled water 7.10 Original buffer solution 1. 35 x 105 M 4.87 4.4 Distilled water with added 6M HCI 1.74 0.01819M Buffer solution with 4.41 6.7x 10 5 M added 6M HCI Distilled water with added 6M NaOH 12.26 5.50 x 10 M Buffer solution with 5.12 7.586 15°M added 6M NaOH ow to calculate [H3O*] and pH?
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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how to find calculated pH？
![Videos: buffer, distilled water and hydrochloric acid https://www.youtube.com/watch?
y=uk6yK2bnCRc
bion onte bbs uov node ler
How buffers are prepared and how they work. https://www.youtube.com/watch?y=4-
C9uz5VXDc
15tlud ors to Hlq
9sd to hion boteoonoo o noitibba R
During the experiment you will determine the capacity of the newly prepared buffer. Buffer
solutions resist a change in pH, and the hydronium ion concentration of a buffer solution therefore
does not change significantly when a small amount of strong acid or strong base is added.
Table 4. Observing pH Changes in Water and Buffer Solutions upon Addition of HCl and NaOH
Solutions.
Solution
Measured pH
Theoretical [H30ʻ] |Calculated pH
Distilled water
7. 10
Original buffer
solution
1.35 x 105 M
4.87
4.4
Distilled water with
added 6M HCI
1.74
0.01819M
Buffer solution with
added 6M HCI
4.41
6.7x 10-5 M
Distilled water with
added 6M NaOH
12.26
13
5.50 x
Buffer solution with
added 6M NaOH
7.586 10°M
ら.12
How to calculate [H3O"] and pH?
Distilled water: We know that Kw= [H3O*[OH] = Ix10-14 , so now you can solve for [H3O*]
pH= -log[1x10-7], solve for pH
Original buffer: You have already solved this question in Table 2.
Distilled water with 6M HCl: You will need to do the dilution part first. Remember, since you added
0.5 mL of 6M HCl to 25.0 mL of distilled water, the total volume is different now. So, what is Ma
25.5mL.
of H3O now? Solve for [H3O*], and find pH
Your text book has a great explanation as well: EXAMPLE 14.20, part a and b.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F9b8e69e2-8fc5-448b-b35a-28a361004eed%2F7dd0z5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Videos: buffer, distilled water and hydrochloric acid https://www.youtube.com/watch?
y=uk6yK2bnCRc
bion onte bbs uov node ler
How buffers are prepared and how they work. https://www.youtube.com/watch?y=4-
C9uz5VXDc
15tlud ors to Hlq
9sd to hion boteoonoo o noitibba R
During the experiment you will determine the capacity of the newly prepared buffer. Buffer
solutions resist a change in pH, and the hydronium ion concentration of a buffer solution therefore
does not change significantly when a small amount of strong acid or strong base is added.
Table 4. Observing pH Changes in Water and Buffer Solutions upon Addition of HCl and NaOH
Solutions.
Solution
Measured pH
Theoretical [H30ʻ] |Calculated pH
Distilled water
7. 10
Original buffer
solution
1.35 x 105 M
4.87
4.4
Distilled water with
added 6M HCI
1.74
0.01819M
Buffer solution with
added 6M HCI
4.41
6.7x 10-5 M
Distilled water with
added 6M NaOH
12.26
13
5.50 x
Buffer solution with
added 6M NaOH
7.586 10°M
ら.12
How to calculate [H3O"] and pH?
Distilled water: We know that Kw= [H3O*[OH] = Ix10-14 , so now you can solve for [H3O*]
pH= -log[1x10-7], solve for pH
Original buffer: You have already solved this question in Table 2.
Distilled water with 6M HCl: You will need to do the dilution part first. Remember, since you added
0.5 mL of 6M HCl to 25.0 mL of distilled water, the total volume is different now. So, what is Ma
25.5mL.
of H3O now? Solve for [H3O*], and find pH
Your text book has a great explanation as well: EXAMPLE 14.20, part a and b.
![For acetic acid, at equilibrium:
der eyohvev te l
Ka= [H3O*][CH;CO2¯V [CH3 CO2 H] =1.8×10-5
In this case, you know Ka value. This is something you would look up in any chemistry book. Set
up your calculations:ollot ntu oeadibios sd lw nouolos les bio he o le
[H3O'][CH;CO2V[CH3 CO2 H] =1.8×105,
noitule tse laiusn s mdt or n liw oed anone a bee bioe snoneA
We know that [H3O*]=[CH3CO2 ], because it's a weak acid and the percent of the acetic acid
molecules that dissociate is so small, it is usually satisfactory to assume the initial acid
concentration equals the equilibrium concentration of the undissociated weak acid. Also, it was
given that [CH3CO2H] 1.0×10"' M
of znoo o oq di orfi to zoiomoo
An2 boteluda iud o sibcs
Now you can solve for [H3O*]:
1- [ CH3COOH]= 1.0xh5!
1. [H3O']/(1.0×10"1 M)=1.8×10-5
2. [H3Oʻ]=0.0013
3. pH=-log[H30]
tinc to s
e hod 4. pH=2.9
DE
CHCOOH.
Table 2. Preparing acetic acid solutions and determining pH
expected.
Concentration
Measured
Theoretical pH
Calculated Ka of acetic
Literature Ka of
of acetic acid,
pH
acid based on measured pH acetic acid
M
0.IM
2.86
2.9
0.5/M
3.20
3.4
3.980 x 10
0.00IM
3.70
-5
3.980 X 10
1.8 x 10-5
3.9
0.000IM
4.17
나.나
4.57x0
PH haip ha
fnd M.
G.M x 8 7
(M)
1. PH = 2.86 -ly l0.003) ICH3COOH ]=o.lM
CHAO 1
ka:
CCH LOO1+3
2. PH=3.2
E CH 2 CoU4 1 =0.0IM
-3.2
=0.006309
0.30363092
Ka :
= 3.980 x 10
-5
'0. 01
3, pH : 3.7o
CH 0J= 10
CHSCOOHI= 0.00IM oisinnd Hq gaiened
3.70 o
0-00019952
0.001
CCH&COOH1:0-001
ka =
-5
: 3.980 x 10
4. PIA = 4.17
CH20"J: 10: 6.7608x105
-4.77
ka - (64608 x10%
4-57 x5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F9b8e69e2-8fc5-448b-b35a-28a361004eed%2F6h33alm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For acetic acid, at equilibrium:
der eyohvev te l
Ka= [H3O*][CH;CO2¯V [CH3 CO2 H] =1.8×10-5
In this case, you know Ka value. This is something you would look up in any chemistry book. Set
up your calculations:ollot ntu oeadibios sd lw nouolos les bio he o le
[H3O'][CH;CO2V[CH3 CO2 H] =1.8×105,
noitule tse laiusn s mdt or n liw oed anone a bee bioe snoneA
We know that [H3O*]=[CH3CO2 ], because it's a weak acid and the percent of the acetic acid
molecules that dissociate is so small, it is usually satisfactory to assume the initial acid
concentration equals the equilibrium concentration of the undissociated weak acid. Also, it was
given that [CH3CO2H] 1.0×10"' M
of znoo o oq di orfi to zoiomoo
An2 boteluda iud o sibcs
Now you can solve for [H3O*]:
1- [ CH3COOH]= 1.0xh5!
1. [H3O']/(1.0×10"1 M)=1.8×10-5
2. [H3Oʻ]=0.0013
3. pH=-log[H30]
tinc to s
e hod 4. pH=2.9
DE
CHCOOH.
Table 2. Preparing acetic acid solutions and determining pH
expected.
Concentration
Measured
Theoretical pH
Calculated Ka of acetic
Literature Ka of
of acetic acid,
pH
acid based on measured pH acetic acid
M
0.IM
2.86
2.9
0.5/M
3.20
3.4
3.980 x 10
0.00IM
3.70
-5
3.980 X 10
1.8 x 10-5
3.9
0.000IM
4.17
나.나
4.57x0
PH haip ha
fnd M.
G.M x 8 7
(M)
1. PH = 2.86 -ly l0.003) ICH3COOH ]=o.lM
CHAO 1
ka:
CCH LOO1+3
2. PH=3.2
E CH 2 CoU4 1 =0.0IM
-3.2
=0.006309
0.30363092
Ka :
= 3.980 x 10
-5
'0. 01
3, pH : 3.7o
CH 0J= 10
CHSCOOHI= 0.00IM oisinnd Hq gaiened
3.70 o
0-00019952
0.001
CCH&COOH1:0-001
ka =
-5
: 3.980 x 10
4. PIA = 4.17
CH20"J: 10: 6.7608x105
-4.77
ka - (64608 x10%
4-57 x5
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