How do you find the corresponding values K of an IndicatorReport sheetNameTimeA1Rmoles= m XVNA = moles acetate present in solution B 3.2X10て216.4%3Dmole Ac6.16molX2.X10-3Molarity HAc in Solution AM HAcN.HACMolesmL SolnAbsor-AbanceNHACNHAC[HIn]HIn]*****Addedat ANACPresent0.06.31 A,=6.821%3DA =0 37|.6X1D1.6X10-40.51.0てb69-7 10-6.0534.754.944-6391 は212.0470A =0•3443.2 X10-4%3D4.50 A-0-307 48 X16153.00.176 4-676 1:69 6:227 4-73A =0.252 6-4YIos454.00.3015.025%3D1 mLHCIAverage pKHAC4-69Corresponding KHAe181**See Equation (11)[HINAverage pKHlnCorresponding KHlnHIn*See Equation (5)*** See Equation (4)Pe= pAj log NAACShow your calculations on the following page.サ- 94%3D%3D

To determine the Corresponding value of KHAc & kHIN

Answered: K of an Indicator ReporT sheet Name

Science

Chemistry

K of an Indicator ReporT sheet Name

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter3: Statistical Tests With Excel

Section: Chapter Questions

Problem 6P

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Question

How do you find the corresponding values

K of an Indicator
Report sheet
Name
Time
A1
R
moles= m XV
NA = moles acetate present in solution B 3.2X10
て216.4
%3D
mole Ac
6.16molX2.X10-3
Molarity HAc in Solution A
M HAc
N.
HAC
Moles
mL Soln
Absor-
A
bance
NHAC
NHAC
[HIn]
HIn]
**
***
Added
at A
NAC
Present
0.0
6.31 A,=6.821
%3D
A =0 37
|.6X1D
1.6X10-4
0.5
1.0
てb69-7 10-
6.0534.75
4.94
4-6391 は21
2.0
470
A =0•3443.2 X10-4
%3D
4.50 A-0-307 48 X16
15
3.0
0.176 4-676 1:69 6:227 4-73
A =0.252 6-4YIo
s45
4.0
0.301
5.0
25
%3D
1 mL
HCI
Average pKHAC
4-69
Corresponding KHAe
181
**See Equation (11)
[HIN
Average pKHln
Corresponding KHln
HIn
*See Equation (5)
*** See Equation (4)
Pe= pAj log NAAC
Show your calculations on the following page.
サ- 94
%3D
%3D

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