based on the equilibrium constant (6.62 * 10^-8) does the equilibrium for equation 4.5( see image 1st image) lay to the right or to the left? how would it shift if base was added to solution 2

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based on the equilibrium constant (6.62 * 10^-8) does the equilibrium for equation 4.5( see image 1st image) lay to the right or to the left? how would it shift if base was added to solution 2 (see second image for rest of the data) 

Квтв
=
*8 bos it and to dige
[B2-][H3O+]
[HB-1
Transcribed Image Text:Квтв = *8 bos it and to dige [B2-][H3O+] [HB-1
Solutions: neqx3
21
219ilstad e 8.45
pH of solution
Josri2 trogen 12209
Concentration of H3O+
Absorbance at max(for B²-)
Concentration of B2- in solution
Concentration of HB in solution
K values
Average K value
A
sob sua
Deviation
EP
teapt
1
Standard deviation cod
8.45 7026
N/A
2
N/A
-a
3.55.10 " 5. S0x1080 1.15 2107 1.78 X100? 263 x10-5
0.774
2.5x10-5
N/A
3
4 5
7026 6.94 6.75 4.58
0.406 0.303 0.205 -0.004
1.31x10³ 9.74×10€ 6.62x10-6 N/A
1.19×10.5 0.000025
1,52x10 1,848/05
160 vehead t01 8520
6.06 xlo 7.39x10 6.41x10
6162×108
8 N/A)
a
S.557x10 7.67 5x16 2.118x10 N/A
WEJ
almanachoad/
9
6.87 x 10.2 bozsd
Transcribed Image Text:Solutions: neqx3 21 219ilstad e 8.45 pH of solution Josri2 trogen 12209 Concentration of H3O+ Absorbance at max(for B²-) Concentration of B2- in solution Concentration of HB in solution K values Average K value A sob sua Deviation EP teapt 1 Standard deviation cod 8.45 7026 N/A 2 N/A -a 3.55.10 " 5. S0x1080 1.15 2107 1.78 X100? 263 x10-5 0.774 2.5x10-5 N/A 3 4 5 7026 6.94 6.75 4.58 0.406 0.303 0.205 -0.004 1.31x10³ 9.74×10€ 6.62x10-6 N/A 1.19×10.5 0.000025 1,52x10 1,848/05 160 vehead t01 8520 6.06 xlo 7.39x10 6.41x10 6162×108 8 N/A) a S.557x10 7.67 5x16 2.118x10 N/A WEJ almanachoad/ 9 6.87 x 10.2 bozsd
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