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Ionic equilibrium

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).

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Ionic equilibrium

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  1. Ionic equilibrium Lec.10

  2. Recall Where : C =1/V weak electrolyte Very weak electrolyte Degree of Ionization

  3. The Ion Product of Water ( Very weak electrolyte)

  4. 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

  5. Example (2): What is the pH value of: • Pure water. • HCl (0.01 mole/L). • NaOH (0.001 mole/L).

  6. 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

  7. 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

  8. 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.

  9. Example (5): If the pH value of 0.1 mole/L acetic acid is 2.872, calculate the ionization constant of the acid. Answer C=0.1 pH= 2.872 K=? pH= -log [H+] 2.872 = -log [H+] [H+]= 1.353x10-3

  10. 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

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