Effect Of Addition Of Strong/Weak Acids And Bases - Ionic Equilibrium | Physical Chemistry PDF Download

Effect of addition of strong electrolyte in weak electrolyte:

• Adding strong electrolyte which is non reacting

• Adding strong electrolyte which is reacting

• Adding strong electrolyte which is having common ion

Case I: Adding  non reacting electrolyte [CH₃CooH + NaCl]

(0.1 M NaCl, volume V)+(0.1MCH₃CooH, Volume V)

Now, concentration of 

Use 0.05 for further calculation not 0.1, because we did mixing which leads to changes in concentration.

Case II: Adding strong electrolyte (reacting) [CH₃OOH + AgNO₃]

                        CH₃COOH            CH₃COO^–    + H⁺

Let at eq.              10                                 5                5

New eq.               10-x                              4+x             5+x

                        AgNO₃                    →          Ag⁺ + NO₃^–

                        1                                           -           -

                        –                                           1          1

                        CH₃COO^– + Ag⁺        →        white ppt.

Kα =

III.  Adding strong electrolyte having common ion[ CH₃COOH + HCl ]

            CH₃COOH           CH₃COO^– + H⁺

            C₁                                    –               –

Eq.       C₁ – C₁α                      C₁α      C₁α

Strong electrolyte                    HCl →  H⁺  + Cl^–

                                                C₂          –       –

                                                –            C₂    C₂

Here added species HCl dissociated to give common ion H⁺.

Because of excess concentration of H⁺, equilibrium shift, to backward direction. Thus,

CH₃COOH                 CH₃COO^– + H⁺

      Equation                           C₁ – C₁α¹                     C₁α¹              C₁α¹

Here    α¹< α

Because equilibrium is shifted to backward direction

Considering α¹<<< 1

C₁α¹<<< C₂

Hence C₁α¹ + C₂≈C₂

So, Kα =

Note:

Here C₂ is the concentration of common ion Not the concentration of acid added.

Example: Calculate the degree of dissociation and pH of the solution

1. If 0.1 m aqueous solution of ammonia is present.

2. If 0.1 m aqueous solution of ammonia is present in presence of 0.05 m Ca(OH)₂, Kb = 10^–5

            NH₄OH                  NH₄⁺ + OH^–

            C - Cα                        Cα       Cα

           α <<< 1

            α = 10^–2

The value of a is less than 0.05. hence neglection of a is correct.

[OH^–] = Ca = 10^–1 × 10^–2 = 10^–3

P_OH = – log [OH^–] = –log (10^–3)]

P_OH = 3

pH = 14–3 = 11

pH = 11

b)    NH₄OH                   NH₄⁺ + OH^–

     C - Cα¹                        Cα¹       Cα¹

    Cα(OH)₂      →              Cα²⁺ + 2 OH^–         

    0.05                              -                -

    -                                   0.05         0.1

Kb =

            After neglection, 10^–5 = (Cα¹ + 0.1) 1

                                    10^–5 = 0.1 × 1

                                    α¹ = 10^–4             

            [OH^–] = Cα¹ + C₂ = 10^–5 + 10^–1 = 10^–1         (10^–5 is neglected as compared to 10^–1)

            [OH^–] = 0.1

            [H⁺] [OH^–] = 10^–14

            [H⁺] =

            pH = 13

Addition of weak electrolyte on weak electrolyte

H              H⁺ + A^–             Kα₁   =        [C₁ – x ≈ C₁]

C₁                    –          –                      Kα₁C₁ = (x+y)(x)

C₁ – x              x+y      x              

HB           H⁺+ B^–                            Kα₂  =

C₂                    –        –

C₂ – x              x+y      y                       Kα₂ C = (x+y) (x)

Add (1) and (2)

Remember, this Formula is after neglection. We have considered that x and y are neglected as compared to C₁

Example: Acetic acid with Kα = 10^–5 is present in an aqueous solution. Calculate the pH of solution if

a)         When only 0.1M acetic acid is present

b)         When only 0.1 Molar acetic acid is present along 0.05 m H₂SO₄.

c)         When 1L acetic acid (0.1 M) is mixed with 1L, 0.1 M HCl

Solution:

a)    

            CH₃COOH       CH₃COO^–   + H₂

            C - Cα                         Cα               Cα

            [H⁺] = Cα

            = 10^–1 x 10^–2 = 10^–3

pH = – log [H⁺] = 3

pH = 3

b)   H₂SO₄→ 2H⁺ + SO₄^2–

            0.05            –          –

            –               0.1      0.05

            CH₃COOH          CH₃COO^–   + H⁺

            C - Cα¹                        Cα¹               Cα¹

Ka = 0.1α¹

α¹ = 10^–4

[H⁺] = Cα¹ + 0.1 = 10^–1 × 10^–4 + 0.1

[H⁺] = 10^–5 + 10^–1

[H⁺] = 10^–1 (we have neglected terms that are less than ten times)

pH = 1

c)         Here, first you have to calculate the concentration of species after mixing

            Concentration of CH₃COOH =   = 0.05 M

            Concentration of HCl = 0.05 M

            CH₃COOH          CH₃COO^– + H⁺

            C - Cα¹                        Cα¹              Cα¹

            HCl     →        H⁺    +   Cl^–

            0.05                 –              –

            -                       0.05     0.05

After considering neglection.

K_α = = 2 X 10^-4

[H⁺] = 0.05 + Cα¹

            [H⁺] = 0.05

            pH = - log (0.05)

            pH = 1.3

Example: 0.1 M NaOH, 200 ml is mixed with 0.1m, 100 ml Ca(OH)₂. Find the pH of the solution. 

Solution: 0.1 M NaOH, 200 ml + 0.1 M, 100 ml Ca(OH)₂

For NaOH = n OH^– = 0.1 × 200 = 20 mmol

Total no. of milimoles of OH^– = 40 mmol

[OH^–] =

P_OH =  

pH = 13.12

Effect of addition of strong Acid on strong base:

When a strong acid is mixed with strong base,

An acid – base Reaction takes place, until the limiting reagent is finished.

For ex: When 0.1M, 100 ml HCl is mixed with 0.1 M, 50 ml NaOH.

It’s obvious the fastest reaction i.e., the Acid – base reaction takes place firstly.

                        HCl + NaOH→NaCl + H₂O

Milimoles        10          5       

                        5          -                 5

(n_H⁺) left = 5 milimoles

Example: 0.1 M, 100 ml HCl is mixed with 0.1M, 100 ml (Ca(OH)₂. Find the pH of the solution.

Solution: As the question itself revealing. When an acid and base is mixed, Firstly acid - Base Reaction takes place.

2 HCl + Ca(OH)₂→ CaCl₂ + H₂O

Milimoles of Ca(OH)₂ = 0.1 × 100 = 10 mmol

2 HCl + Ca(OH)₂→ CaCl₂ + H₂O

10            10

–               5                 5

Now we have only Ca(OH)₂ in the solution. It will completely dissociate to give Ca²⁺ and 2OH^–

Ca(OH)₂→ Ca²⁺ + 2OH^–

5                    -             -

-                   5         10

(OH^–) =  = .05

pOH = 1.3

Hence pH = 12.7