For acetic acid, at equilibrium: Ka- [H3O*][CH;CO2 V [CH, CO2 H] =1.8×10"s In this case, you know Ka value. This is something you would look up in any chemistry book. Set up your calculations: [H3O°][CH3CO2 V[CH3 CO2 H] =1.8×10-5, noiulos iie lee 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 valmoo Now you can solve for [H3O*]: 1-[ CH3COOH]=1.0xW' 1. [H3O*]³/(1.0×10' M)=1.8×105 2. [H3O']=0.0013 3. pH=-log[H3Oʻ] iee lo sm 4. pH=2.9 CH COOH. Table 2. Preparing acetic acid solutions and determining pH expected. Literature Ka of Measured Theoretical pH Calculated Ka of acetic Concentration acid based on measured pH acetic acid of acetic acid, M pH 0.IM 2.86 -5 -5 3.980 x 10 0.5/M 8.20 -5 3.980 X 10 ç. 01 x 87 0.00/M 3.70 5 4.17 4.57xp 0.000|M 1. PH = 2.86 -ly l0.0000) I CH2 COUH ]=0.IM [H30 ]= 15-0.00138. CH0 ka: CCH CO0I+] 256 0.1
For acetic acid, at equilibrium: Ka- [H3O*][CH;CO2 V [CH, CO2 H] =1.8×10"s In this case, you know Ka value. This is something you would look up in any chemistry book. Set up your calculations: [H3O°][CH3CO2 V[CH3 CO2 H] =1.8×10-5, noiulos iie lee 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 valmoo Now you can solve for [H3O*]: 1-[ CH3COOH]=1.0xW' 1. [H3O*]³/(1.0×10' M)=1.8×105 2. [H3O']=0.0013 3. pH=-log[H3Oʻ] iee lo sm 4. pH=2.9 CH COOH. Table 2. Preparing acetic acid solutions and determining pH expected. Literature Ka of Measured Theoretical pH Calculated Ka of acetic Concentration acid based on measured pH acetic acid of acetic acid, M pH 0.IM 2.86 -5 -5 3.980 x 10 0.5/M 8.20 -5 3.980 X 10 ç. 01 x 87 0.00/M 3.70 5 4.17 4.57xp 0.000|M 1. PH = 2.86 -ly l0.0000) I CH2 COUH ]=0.IM [H30 ]= 15-0.00138. CH0 ka: CCH CO0I+] 256 0.1
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 calculate for theoretical pH？？
![MacBook Air
14
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4
3
5
6
8
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delete
For acetic acid, at equilibrium:
Ka= [H3O*][CH3CO2 V [CH3 CO2 H] =1.8×10¯s
In this case, you know Ka value. This is something you would look up in any chemistry book. Set
up your calculations:
[H3O*][CH3CO2/[CH3 CO2 H] =1.8×10¯³,
noihulos ilne leiusn s ol or ed anon
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
bios
eum 2st o o
Now you can solve for [H3O*]:
sldul
ohdoa ilo ztorioto H
1- [ CH3COOH]=1-0x W'
1. [H3O*]/(1.0×10"' M)=1.8×10-s
2. [H3O*]=0.0013
3. pH=-log[H3O"] w1sd
4. pH=2.9
ansiqx
bo i
ise lo sme
H sdoibins)
CH COOH
Table 2. Preparing acetic acid solutions and determining pH
expected.
Literature Ka of
Calculated Ka of acetic
acid based on measured pH acetic acid
Concentration
Measured
Theoretical pH
of acetic acid,
pH
0.IM
2.86
G.O! X Gcb'l
--
-5
3.980 x 10
0.5/M
3.80
-5
3.980 X 10
s-01 X 8%
0.00IM
3.70
4.17
4.57x5
0.000|M
-leyl0.000) ICH2COUH =0.IM
1. PH= 2.86
CHO
ka=
CCH COO1+]
2blow om
-5
odt g
U-1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F3d642c89-6e87-42bc-b24a-ae1cda03948a%2Fis283o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:MacBook Air
14
2$
4
3
5
6
8
T
Y
U
delete
For acetic acid, at equilibrium:
Ka= [H3O*][CH3CO2 V [CH3 CO2 H] =1.8×10¯s
In this case, you know Ka value. This is something you would look up in any chemistry book. Set
up your calculations:
[H3O*][CH3CO2/[CH3 CO2 H] =1.8×10¯³,
noihulos ilne leiusn s ol or ed anon
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
bios
eum 2st o o
Now you can solve for [H3O*]:
sldul
ohdoa ilo ztorioto H
1- [ CH3COOH]=1-0x W'
1. [H3O*]/(1.0×10"' M)=1.8×10-s
2. [H3O*]=0.0013
3. pH=-log[H3O"] w1sd
4. pH=2.9
ansiqx
bo i
ise lo sme
H sdoibins)
CH COOH
Table 2. Preparing acetic acid solutions and determining pH
expected.
Literature Ka of
Calculated Ka of acetic
acid based on measured pH acetic acid
Concentration
Measured
Theoretical pH
of acetic acid,
pH
0.IM
2.86
G.O! X Gcb'l
--
-5
3.980 x 10
0.5/M
3.80
-5
3.980 X 10
s-01 X 8%
0.00IM
3.70
4.17
4.57x5
0.000|M
-leyl0.000) ICH2COUH =0.IM
1. PH= 2.86
CHO
ka=
CCH COO1+]
2blow om
-5
odt g
U-1
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