Question

Part A Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.8×10−5. Part B Find the percent dissociation of this solution. Part C Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.4×10−3. Part D Find the percent dissociation of this solution. Part E Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 0.15. Part F Find the percent dissociation of this solution. Thank you!

Answer 1

PART A & B:

Let the monoprotic weak acid is represented as HA

Make ICE table.

            HA     +      H2O      ---->      A-      +     H3O+
IC:      0.12                                          0                  0
C:         -x                                          +x                +x
EC:   0.12-x                                        x                   x

Ka = [A^-] [H₃O⁺] / [HA]

1.8 x 10^-5 = (x) (x) / (0.12 – x)

1.8 x 10^-5 = x² / (0.12)          ; Since Ka is very small

x² = (1.8 x 10^-5) (0.12)

x² = 2.16 x 10^-6

x = 1.5 x 10^-3 = 0.0015

So,

[H₃O⁺] = x = 1.5 x 10^-3

pH = -log [H₃O⁺]

     = - log(1.5 x 10^-3)

     = 2.82

Percent dissociation = (x/initial concentration)*100%

                                   =(0.0015 / 0.120) x100 %

                                  = 1.25 %

PART E & F:

Let the monoprotic weak acid is represented as HA

Make ICE table.

            HA     +      H2O      ---->      A-      +     H3O+
IC:      0.12                                          0                  0
C:         -x                                          +x                +x
EC:   0.12-x                                        x                   x

Ka = [A^-] [H₃O⁺] / [HA]

0.15 = (x) (x) / (0.12 – x)

0.15 = x² / (0.12 - x)

x² = (0.12 - x) (0.15)

x² = 0.018 – 0.15x

x² + 0.15x - 0.018 = 0

Solving the quadratic equation, we get;

x = 0.079 and x = -0.229

Discarding negative value of x. So, x = 0.079

[H₃O⁺] = x = 0.079

pH = -log [H₃O⁺]

     = - log(0.079)

     = 1.10

Percent dissociation = (x/initial concentration)*100%

                                   =(0.079 / 0.120) x100 %

                                  = 65.83 %

PART C & D:

Let the monoprotic weak acid is represented as HA

Make ICE table.

            HA     +      H2O      ---->      A-      +     H3O+
IC:      0.12                                          0                  0
C:         -x                                          +x                +x
EC:   0.12-x                                        x                   x

Ka = [A^-] [H₃O⁺] / [HA]

1.4 x 10^-3 = (x) (x) / (0.12 – x)

0.0014 = x² / (0.12 - x)

x² = (0.12 - x) (0.0014)

x² = 0.000168 – 0.0014x

x² + 0.0014x - 0.000168 = 0

Solving the quadratic equation, we get;

x = 0.012 and x = -0.014

Discarding negative value of x. So, x = 0.012

[H₃O⁺] = x = 0.012

pH = -log [H₃O⁺]

     = - log(0.012)

     = 1.92

Percent dissociation = (x/initial concentration)*100%

                                   = (0.012 / 0.120) x100 %

                                  = 10 %