0.008moles acetate present in solution BO,16mole AcAcMolarity HAc in Solution AМ НАСNMolesmL SolnAbsor-HAC[HinIn ][Hin]NHAClogpKin ]NHACpKAbancePHlogHinHACHC Н,о,PresentNACNACAddedat AmaxXXXA,=0.5640.06.71-0.4964.170.00 8614.68.020.00&160.3191.04.67 A =0.3190.6543.7260.01764. 1OA =0.222 6 ,00 8324,380.2222.0l.040.772 3.Y18C.1690.0253 22A 0 ol69 000 212,063.04.19-0.863 3.2170.1370.0334 1I7L.084.08 A-0.1370.00264.00.9243.0660.1196.0414 14403A =0. 119 000 883.995.05.0+2.54XXA =0.153X1 mLНCI4.29Corresponding KAe.5X10HAcAverage pKHAc3.02x10-9Corresponding K3.S2HInAverage pKHIn***See Equation (4)**See Equation (11)See Equation (5)39

Question
Asked Oct 29, 2019

It was assumed:

a) Before any acetic acid was added, essentially all the indicator was in the In- form. 

b) After the HCl was added, essentially all the indicator was in the HIn form. 

Justify these assumptions using the measured pH of these solutions and your experimental value for pKHIn to determine:

1) The fraction fo the indicator in the HIn form in solution B before any of solution A was added

2) The fraction of the indicator in the In- form after the HCl was added. 

 

0.008
moles acetate present in solution B
O,16
mole Ac
Ac
Molarity HAc in Solution A
М НАС
N
Moles
mL Soln
Absor-
HAC
[Hin
In ]
[Hin]
NHAC
log
pK
in ]
NHAC
pK
A
bance
PH
log
Hin
HAC
HC Н,о,
Present
NAC
NAC
Added
at A
max
X
X
X
A,=0.564
0.0
6.71
-0.4964.17
0.00 8614.68
.02
0.00&16
0.319
1.0
4.67 A =0.319
0.654
3.726
0.01764. 1O
A =0.222 6 ,00 832
4,38
0.222
2.0
l.04
0.772 3.Y18
C.169
0.0253 22
A 0 ol69 000 212,06
3.0
4.19
-0.863 3.217
0.137
0.0334 1I7
L.08
4.08 A-0.1370.0026
4.0
0.9243.066
0.119
6.0414 14403
A =0. 119 000 88
3.99
5.0
5.0+
2.54
X
X
A =0.153
X
1 mL
НCI
4.29
Corresponding KAe.5X10
HAc
Average pKHAc
3.02x10-9
Corresponding K
3.S2
HIn
Average pK
HIn
***See Equation (4)
**See Equation (11)
See Equation (5)
39
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Image Transcriptionclose

0.008 moles acetate present in solution B O,16 mole Ac Ac Molarity HAc in Solution A М НАС N Moles mL Soln Absor- HAC [Hin In ] [Hin] NHAC log pK in ] NHAC pK A bance PH log Hin HAC HC Н,о, Present NAC NAC Added at A max X X X A,=0.564 0.0 6.71 -0.4964.17 0.00 8614.68 .02 0.00&16 0.319 1.0 4.67 A =0.319 0.654 3.726 0.01764. 1O A =0.222 6 ,00 832 4,38 0.222 2.0 l.04 0.772 3.Y18 C.169 0.0253 22 A 0 ol69 000 212,06 3.0 4.19 -0.863 3.217 0.137 0.0334 1I7 L.08 4.08 A-0.1370.0026 4.0 0.9243.066 0.119 6.0414 14403 A =0. 119 000 88 3.99 5.0 5.0+ 2.54 X X A =0.153 X 1 mL НCI 4.29 Corresponding KAe.5X10 HAc Average pKHAc 3.02x10-9 Corresponding K 3.S2 HIn Average pK HIn ***See Equation (4) **See Equation (11) See Equation (5) 39

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Expert Answer

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Step 1

Henderson-Hasselbalch equation is shown in equation (1) where pH is for the solution, pKInH is the equilibrium constant for the given indicator, and [In]/[InH] is a fraction of [In] with respect to [InH].

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Step 2

The values for pH and pKInH are substituted in equation (1) to calculate the fraction [InH]/[In]. The fraction [InH]/[In] in the solution, before the addition of solution B, i...

In]
6.71 3.52 log
InH
In]
1067-3.52
InH
InH
103.19
In]
= 6.45 x 10
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In] 6.71 3.52 log InH In] 1067-3.52 InH InH 103.19 In] = 6.45 x 10

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