Help me please to fill the table 2 below with equations and explanation. Write e
ID: 1033004 • Letter: H
Question
Help me please to fill the table 2 below with equations and explanation. Write equation reactions just for hydrolyzing ion with water. Write Ka or Kb expressions for each and calculate Ka and Kb from data.
Table 1. Hydronium and Hydroxide Ion Concentrations
0.10 M Solutions
Bromothymol Blue color
pH (measured)
[H+]
[OH-]
NaCl
Green
6.06
8.71*10^-7
1.15*10^-8
Na3C6H5O7
Blue
8.70
2.00*10^-9
5.01*10^-6
Na2CO3
Blue
11.47
3.39*10^-12
2.95*10^-3
NH4Cl
Green
7.01
9.77*10^-8
1.02*10^-7
NaHSO4
Yellow
1.56
2.75*10^-2
3.63*10^-13
NH4C2H3O2
Green
6.85
1.41*10^-7
7.08*10^-8
Na2SO3
Blue
9.81
1.55*10^-10
6.46*10^-5
Table 2. Hydrolysis of Salts
0.10 M Solutions
Hydrolysis Ions
Spectator Ions
Ka
Kb
NaCl
Na3C6H5O7
Na2CO3
NH4Cl
NaHSO4
NH4C2H3O2
Na2SO3
Thank you!
0.10 M Solutions
Bromothymol Blue color
pH (measured)
[H+]
[OH-]
NaCl
Green
6.06
8.71*10^-7
1.15*10^-8
Na3C6H5O7
Blue
8.70
2.00*10^-9
5.01*10^-6
Na2CO3
Blue
11.47
3.39*10^-12
2.95*10^-3
NH4Cl
Green
7.01
9.77*10^-8
1.02*10^-7
NaHSO4
Yellow
1.56
2.75*10^-2
3.63*10^-13
NH4C2H3O2
Green
6.85
1.41*10^-7
7.08*10^-8
Na2SO3
Blue
9.81
1.55*10^-10
6.46*10^-5
Explanation / Answer
Table 2
0.1 M NaCl
H+ + OH- <==> H2O
spectator ion = Na+, Cl-
Ka = [H+]^2 = (8.71 x 10^-7)^2 = 7.6 x 10^-13
Kb = Kw/[H+] = 1 x 10^-14/7.6 x 10^-13 = 0.0132
--
0.1 M Na3C6H5O7
C6H5O7^3- + H2O <==> HC6H5O7^2- + OH-
Ka = Kw/Kb = 1 x 10^-14/2.51 x 10^-10 = 3.98 x 10^-5
Kb = [HC6H5O7^2-][OH-]/[C6H5O7^3-] = (5.01 x 10^-6)^2/0.1 = 2.51 x 10^-10
--
0.1 M Na2CO3
CO3^2- + H2O <==> HCO3- + OH-
Ka = Kw/Kb = 1 x 10^-14/8.97 x 10^-5 = 1.14 x 10^-10
Kb = [HCO3-][OH-]/[CO3^2-] = (2.95 x 10^-3)^2/(0.1 - 0.00295) = 8.97 x 10^-5
--
0.1 M NH4Cl
NH4+ + H2O <==> NH3 + H3O+
Ka = [NH3][H3O+]/[NH4+] = (9.77 x 10^-8)^2/0.1 = 9.5 x 10^-14
Kb = Kw/Ka = 1 x 10^-14/9.5 x 10^-14 = 0.105
--
0.1 M NaHSO4
HSO4- + H2O <==> H2SO4 + OH-
Ka = Kw/Kb = (2.75 x 10^-2)^2/(0.1-2.75 x 10^-2) = 0.0104
Kb = [H2SO4][OH-]/[HSO4-] = (3.63 x 10^-13)^2/0.1 = 1.32 x 10^-24
--
0.1 M NH4C2H3O2
NH4+ H2O <==> NH3 + H3O+
C2H3O2- + H2O <==> HC2H3O2 + OH-
Ka = [NH3][H3O+]/[NH4+] = (1.41 x 10^-7)^2/0.1 = 2 x 10^-13
Kb = [HC2H3O2][OH-]/[C2H3O2-] = (7.08 x 10^-8)^2/0.1 = 5 x 10^-14
--
0.1 M Na2SO3
SO3^2- + H2O <==> HSO3- + OH-
Ka = Kw/Kb = 1 x 10^-14/4.2 x 10^-8 = 2.4 x 10^-7
Kb = [HSO3-][OH-]/[SO3^2-] = (6.46 x 10^-5)^2/0.1 = 4.2 x 10^-8
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