Which of the ions that you studied underwent hydrolysis? Explain whether or not

Question

Which of the ions that you studied underwent hydrolysis? Explain whether or not your results were as expected (i.e. acidic/basic/neutral solution).

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

# of equetion

0.10 M Solutions

Hydrolysis Ions

Spectator Ions

Ka

Kb

7.1

NaCl

-

Na+, Cl–

7.6 x 10^-13

0.0132

7.2

Na3C6H5O7

C6H5O73–

Na+

3.98 x 10^-5

2.51 x 10^-10

7.3

Na2CO3

CO32–

Na+

1.14 x 10^-10

8.97 x 10^-5

7.4

NH4Cl

NH4+

Cl–

9.5 x 10^-14

0.105

7.5

NaHSO4

-

Na+, SO42–

0.0104

1.32 x 10^-24

7.6

NH4C2H3O2

NH4+ , CH3–COO–

-

2 x 10^-13

5 x 10^-14

7.7

Na2SO3

SO32–

Na+

2.4 x 10^-7

4.2 x 10^-8

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

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

Saltsof weak acid or weak base goes hydrolysis in water.

For salt of weak acid,M=cation ion,X=anion

MaXb <-->X^a- +M^b+

X^a- +H2O <--->HX^(a+1) +OH- (hydrolysis) Kb=[OH-][HX^(a+1)]/[X^a-]

For salt of weak base,M=cation,X=anion

M^b+ +H2O <--> (MOH^(b-1)) +H3O+ (hydrolysis)

ka=[H3O+][(MOH^(b-1))]/[M^b+]

hydrolysis undergoneby the following ions:

1)

0.1 M Na3C6H5O7

C6H5O7^3- + H2O <==> HC6H5O7^2- + OH-

2)

0.1 M Na2CO3

CO3^2- + H2O <==> HCO3- + OH-

3) NH4Cl

NH4+ + H2O <==> NH3 + H3O+

4) NaHSO4

HSO4- + H2O <==> H2SO4 + OH-

5) NH4C2H3O2

NH4+ H2O <==> NH3 + H3O+

6) Na2SO3

SO3^2- + H2O <==> HSO3- + OH-

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