Question

Calculate the Ka of if 5.0 M HF, 2.7% dissociated:

a) Dissociation equilibria:

b) ICE setup:

c) Amount dissociated:

d) Ka:

12. Calculate the [H+], [OH-] and pH of 0.90 M HC2H3O2; Ka= 1.8 x 10-5.

a) Dissociation equilibria:

b) ICE setup:

c) [H+] and [OH-]:

d) pH:

13. Calculate the [H+], [OH-] pOH and pH 5.0 M NH3; Kb= 1.8 x 10^-5

a) Dissociation equilibria:

b) ICE setup

c) [H+] and [OH-]:

d) pOH and pH

Answer 1

Solution

Calculate the Ka of if 5.0 M HF, 2.7% dissociated:

a) Dissociation equilibria:

HF ------- > H+ (aq) + F-(aq)

2.7 % is dissociated therefore the concentration of the H+ = 5.0 M * 2.7 % / 100 % = 0.135 M

b) ICE setup:

          HF ------- > H+ (aq) + F-(aq)

I 5.0 M                  0                0

C         -x                +x             +x

E   5.0 –x               x               x

c) Amount dissociated:

2.7 % is dissociated therefore the concentration of the H+ = 5.0 M * 2.7 % / 100 % = 0.135 M

d) Ka:

Ka = [ H+] [F-] / [HF]

      = [0.135 ][0.135] /[5.0-0.135]

      = 3.75E-3

12. Calculate the [H+], [OH-] and pH of 0.90 M HC2H3O2; Ka= 1.8 x 10-5.

a) Dissociation equilibria:

HC2H3O2 --------- > H+ + C2H3O2 -

b) ICE setup:

          HC2H3O2 --------- > H+ + C2H3O2 –

I          0.90   M                      0                 0

C          -x                              +x               +x

E      0.90 –x                        x                 x

Ka = [H+] [C2H3O2 –] / H[C2H3O2]

1.8E-5 = [x][x] /[0.90 –x]

Since value of the ka is small therefore we can neglect the x from denominator then we get

1.8E-5 = [x][x] /[0.90 ]

1.8E-5 * 0.90 = x^2

1.62E-5 = x^2

Taking square root of both sides we get

4.02E-3=x =[H+]

c) [H+] and [OH-]:

[H+] = 4.02E-3 M

[OH-] = kw / [H+]

           = 1E-14 / 4.02E-3

           = 2.49E-12 M

d) pH:

pH= -log [H+]

pH= -log [ 4.02E-3]

pH= 2.40

13. Calculate the [H+], [OH-] pOH and pH 5.0 M NH3; Kb= 1.8 x 10^-5

a) Dissociation equilibria:

NH3 + H2O ------ > NH4+    + OH-

b) ICE setup

     NH3 + H2O ------ > NH4+    + OH-

I    5.0 M                            0               0

C      -x                             +x                +x

E   5.0 –x                          x                  x

Kb = [NH4+] [OH-] / [NH3]

1.8E-5 = [x][x]/[5.0-x]

Since ka is small we can neglect the value of x from the denominator then we get

1.8E-5 = [x][x]/[5.0]

1.8E-5 * 5.0 = x^2

9E-5 = x^2

Taking square root of both sides we get

9.49E-3 M =x=[OH-]

c) [H+] and [OH-]:

[OH- ] = 9.49E-3 M

[H+] = Kw / [OH-]

[H+] = 1E-14 / 9.49E-3

[H+] = 1.05E-12 M

d) pOH and pH

pOH = -log [OH-]

         = - log [ 9.49E-3]

        = 2.02

pH + pOH= 14

therefore pH= 14 – pOH

                         = 14 – 2.02

                         = 11.98