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# Ionic equilibrium PowerPoint PPT Presentation

Ionic equilibrium. Lec.10. Recall. Where : C =1/V. weak electrolyte. Very weak electrolyte. Degree of Ionization. The Ion Product of Water. ( Very weak electrolyte ). The Hydrogen Ion Exponent:( pH).

Ionic equilibrium

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## Ionic equilibrium

Lec.10

Recall

Where : C =1/V

weak electrolyte

Very weak electrolyte

Degree of Ionization

The Ion Product of Water

( Very weak electrolyte)

### The Hydrogen Ion Exponent:( pH)

• pH is defined as the negative exponent of 10 which gives the hydrogen ion concentration , pH= - log 10[H+]

• pOH = - log [OH-]

• [H+][OH-] = Kw =10-14

• -log [H+]-log[OH-]=-log10-14

• p H +p OH =14

Example (2):

What is the pH value of:

• Pure water.

• HCl (0.01 mole/L).

• NaOH (0.001 mole/L).

Example (3):

The ionization constant of acetic acid at 25 0C is 1.82x10-5.

Calculate the pH of 0.1 mole/ L acid.

Another method of calculating [H+]: According to Ostwald’s dilution law,

pH= -log [H+]= 2.87

### Common Ion Effect

• The ionization of a weak electrolyte is diminished by the addition of a strong electrolyte, which lead to a common ion, acetic acid is largely suppressed by the addition of either sodium acetate or hydrochloric acid(strong electrolyte).

• Cs : the concentration of the strong electrolyte(completely ionized), moles/liter

• C : the original concentration of the weak electrolyte

( including both dissociated and un dissociated molecules)

• α- : the degree of ionization in the presence of the strong electrolyte

Example (4):

The equivalent conductance of acetic acid at infinite dilution is 387 cm3 atm-1 mole-1 at the same temperature but at the dilution of 1 mole in 1000 liters, the equivalent conductance is 55cm3 atm-1 mole-1. Find the percentage ionization of 0.1 mole/L acetic acid solution.

Example (5):

If the pH value of 0.1 mole/L acetic acid is 2.872,

calculate the ionization constant of the acid.

C=0.1pH= 2.872K=?

pH= -log [H+]

2.872 = -log [H+]

[H+]= 1.353x10-3

• Assignment :

• What is the [H]+ conc. of 0.01 N of acetic acid ?

• If 1.64 gm Na-acetate is dissolved in 1 L of 0.01 N of acetic acid , what will be the[H]+ conc.?

• Ka=1.8*10-5, M.wtof Na-acetate =82