Concentration Measured Theoretical pH Calculated Ka of acetic Literature Ka of of acetic acid, pH acid based on measured pH | acetic acid M 2. 86 习2.9 1.905 x 105 0.IM 5-M x 8 7 S-01 x 87 3.20 3.4 3.980 x O -5 0.00IM 3.70 3.9 3.980 X 10 0.000IM 4.17 나.나 4.57xp5 G.M x 8 7
Concentration Measured Theoretical pH Calculated Ka of acetic Literature Ka of of acetic acid, pH acid based on measured pH | acetic acid M 2. 86 习2.9 1.905 x 105 0.IM 5-M x 8 7 S-01 x 87 3.20 3.4 3.980 x O -5 0.00IM 3.70 3.9 3.980 X 10 0.000IM 4.17 나.나 4.57xp5 G.M x 8 7
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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according to the table，how to calculate average Ka？？？
![For acetic acid, at equilibrium:
In this case, you know Ka value. This is something you would look up in any chemistry bOOK. Set
Joibang hao uoY Ha
enoitsiniz
up your calculations:ollt feton o
iw nonulo
[H3O*][CH3COV [CH3 CO2 H]=1.8×10"³,
noliulo ilsa inona nt o s liw oend anone s boe bras noue A
We know that [H3O*]= [CHCO, ], 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, 1t was
given that [CH3CO2H] 1.0×107'M
3.
Now you can solve for [H3o*]:
Teum 2sgrinoo orl lo 20ulevboteludet oiaud 1o oibio
oiloa l ztoro to Hg y E sldnl
1- [ CH3COOH]=1.0xh5'
1. [H3O*]/(1.0×10"' M)=1.8×10-5
2. [H3O*]=0.0013
3. pH=-log[H30']n sd
sle
Hez lo sms
H ad oibos)
e Hod 4. pH=2.9
CH COOH.
Table 2. Preparing acetic acid solutions and determining pH
expected.
Concentration
Measured
Theoretical pH
Calculated Ka of acetic
acid based on measured pH acetic acid
Literature Ka of
of acetic acid,
pH
0.IM
2.86
均 2.9
1.905 x 100
HV
0.5/M
3.20
3.4
-5
3.980 x W
old
0.00IM
3.70
3.어
-5
3.980 X.10
1.8 x 10-5
po ns
0.000IM
4.17
나나
4.57x105
1.8x 5
ed im.
1. PH = 2.86 -ly l0.00130) ICH3COUH ]=0.IM
(M)
ka=
CCH COOI+]
-cw38?g bluow uoy Inominsqe ot gnnuG
-5
2. pH= 3.2
o oton mobo to doso to lom01 C
C CH3COO나 그 "0.oIM
-3.2
=0.0006309
ka :
= 3.980 x 105
"0. 01
3. pH: 3.70
CH 0*J= 10
OH 1a noi
SL CH3COOHI = 0-00IMba 1stewni 2agnad Rq anived
0-00019952
-5
- 3.980 x 10
ka =
%3D
0.001
4. PlH = 4.17
CH20J 10
-4.7
(6.4608 x 105
ka :
= 4-57 x
0.00%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2Faa9dbc41-e981-4357-8dff-5137d972b002%2Fl6kutw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For acetic acid, at equilibrium:
In this case, you know Ka value. This is something you would look up in any chemistry bOOK. Set
Joibang hao uoY Ha
enoitsiniz
up your calculations:ollt feton o
iw nonulo
[H3O*][CH3COV [CH3 CO2 H]=1.8×10"³,
noliulo ilsa inona nt o s liw oend anone s boe bras noue A
We know that [H3O*]= [CHCO, ], 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, 1t was
given that [CH3CO2H] 1.0×107'M
3.
Now you can solve for [H3o*]:
Teum 2sgrinoo orl lo 20ulevboteludet oiaud 1o oibio
oiloa l ztoro to Hg y E sldnl
1- [ CH3COOH]=1.0xh5'
1. [H3O*]/(1.0×10"' M)=1.8×10-5
2. [H3O*]=0.0013
3. pH=-log[H30']n sd
sle
Hez lo sms
H ad oibos)
e Hod 4. pH=2.9
CH COOH.
Table 2. Preparing acetic acid solutions and determining pH
expected.
Concentration
Measured
Theoretical pH
Calculated Ka of acetic
acid based on measured pH acetic acid
Literature Ka of
of acetic acid,
pH
0.IM
2.86
均 2.9
1.905 x 100
HV
0.5/M
3.20
3.4
-5
3.980 x W
old
0.00IM
3.70
3.어
-5
3.980 X.10
1.8 x 10-5
po ns
0.000IM
4.17
나나
4.57x105
1.8x 5
ed im.
1. PH = 2.86 -ly l0.00130) ICH3COUH ]=0.IM
(M)
ka=
CCH COOI+]
-cw38?g bluow uoy Inominsqe ot gnnuG
-5
2. pH= 3.2
o oton mobo to doso to lom01 C
C CH3COO나 그 "0.oIM
-3.2
=0.0006309
ka :
= 3.980 x 105
"0. 01
3. pH: 3.70
CH 0*J= 10
OH 1a noi
SL CH3COOHI = 0-00IMba 1stewni 2agnad Rq anived
0-00019952
-5
- 3.980 x 10
ka =
%3D
0.001
4. PlH = 4.17
CH20J 10
-4.7
(6.4608 x 105
ka :
= 4-57 x
0.00%
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