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lecture 5

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lecture 5

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  1. Course Name: Volumetric and Gravimetric Analytical Chemistry Course Code: 4022133-3

  2. Lecture 5 contents ➢Self-ionization of Water ➢ pH of different Solutions ➢Calculation of pH for different solutions

  3. Self-ionization of Water Self-ionization: is a reaction in which two like molecules react to give ions. - In the case of water, the following equilibrium is established. ) l ( O H ) l ( O H 2 2 + + + + − − + + H O ( aq ) OH ( aq ) 3 - The equilibrium-constant expression for this system is: O H [ K = = + + − − OH ][ ] 3 c 2 [ H O ] 2

  4. Self-ionization of Water -The concentration of ions is extremely small, so the concentration of H2O remains essentially constant. This gives: + + − − 2 = = [ H constant O ] K [ H O OH ][ ] 2 c 3

  5. Self-ionization of Water -Self-ionization is called - We call the equilibrium value for the ion product [H3O+][OH-] the ion-product constant for water, which is written Kw. O H [ K 3 w = = + + − − OH ][ ] - At 25 oC, the value of Kwis 1.0 x 10-14. - Like any equilibrium constant, Kwvaries with temperature.

  6. Self-ionization of Water - Because we often write H3O+as H+, the ion- product constant expression for water can be written: + + − − = = Kw [ H OH ][ ] - Using Kwyou can calculate the concentrations of H+and OH-ions in pure water.

  7. Self-ionization of Water • These ions are produced in equal numbers in pure water, so if we let x = [H+] = [OH-] − − 14= = o   0 . 1 10 x ( ) x )( at 25 C − − − − 14 7 = =   = =   x 0 . 1 10 0 . 1 10 - Thus, the concentrations of H+and OH-in pure water are both 1.0 x 10-7M. - If you add acid or base to water they are no longer equal but the Kwexpression still holds.

  8. Solutions of Strong Acid or Base - By dissolving substances in water, you can alter the concentrations of H+(aq) and OH-(aq). - In a neutral solution, the concentrations of H+(aq) and OH-(aq) are equal, as they are in pure water. - In an acidic solution, the concentration of H+(aq) is greater than that of OH-(aq). - In a basic solution, the concentration of OH-(aq) is greater than that of H+(aq).

  9. Solutions of Strong Acid or Base At 25°C, you observe the following conditions. – In an acidic solution, [H+] > 1.0 x 10-7M. – In a neutral solution, [H+] = 1.0 x 10-7M. – In a basic solution, [H+] < 1.0 x 10-7M.

  10. The pH of a Solution Although you can quantitatively describe the acidity of a solution by its [H+], it is often more convenient to give acidity in terms of pH. pH of a solution -The pH of a solution is defined as the negative logarithm of the molar hydrogen-ion concentration. + + = = − − pH log[ H ]

  11. Calculation of the pH - For a solution in which the hydrogen-ion concentration is 1.0 x 10-3, the pH is: − 3= = −  log( 0 . 1 10 ) . 3 00 pH - Note that the number of decimal places in the pH equals the number of significant figures in the hydrogen-ion concentration.

  12. The pH of a Solution - In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10-7, the pH = 7.00. - For acidic solutions, the hydrogen-ion concentration is greater than 1.0 x 10-7, so the pH is less than 7.00. - Similarly, a basic solution has a pH greater than 7.00.

  13. The pH of a Solution

  14. The pH of a Solution - A measurement of the hydroxide ion concentration, similar to pH, is the pOH. - The pOH of a solution is defined as the negative logarithm of the molar hydroxide-ion concentration. log[ pOH − − = = − − OH ] -Then because Kw= [H+][OH-] = 1.0 x 10-14at 25 oC, you can show that pOH pH + + = = 14 00 .

  15. pH Calculation pH: is the negative logarithmic of the molar concentration of hydronium ion. log[+ − = H pH ] pOH is the negative logarithm of the molar hydroxide- ion concentration. [OH pOH − = log − ] +pOH pH = 14 00 .

  16. A sample of orange juice has a hydrogen-ion concentration of 2.9 x 10-4M. What is the pH? + + = = − − pH log[ H ] − − 4 = = − −   pH pH = = log( 9 . 2 10 ) . 3 54

  17. Example An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3M. What is the pH of the solution? You first calculate the pOH: − − 3= = = = − −   pOH log( 9 . 1 10 ) . 2 72 Then the pH is: pH = = − − = = 14 00 . . 2 72 11 28 .

  18. pH = 1.53

  19. Calculation of pH for different solutions ➢ Strong acid ➢ Strong base ➢ Weak acid ➢ Weak base ➢ Salts

  20. pH of acid and base Weak base, e.g, NH4OH Strong acid, e.g, HCl,HNO3,H2SO4 Weak acid,e.g, CH3COOH Strong base, e.g, NaOH, KOH [base K b ] [OH-] = [base] =[OH-] pH= -log [H+] pOH= -log [OH-] [acid] =[H+] [acid k a ] [H+] = pH= -log [H+] pH= -log [H+]

  21. In case of strong acid - Strong acid is completely ionized, e.g: + H+ Cl- HCl 0.1M 0.1M 0.1M pH= -log [H+] pH= - log [0.1] = 1.0

  22. Example Calculate the pH of 0.001 M HCl

  23. In case of strong base Na++ OH- e.g: Na OH 0.001M 0.001M 0.001M pOH= - log [OH-] pOH = -log 0.001 = 3.0 pH= 14 – pOH = 14 – 3 = 11

  24. Example Calculate the pH of a 0.020 M NaOH solution

  25. In case of weak acid HA(aq) + H2O(l) H3O+(aq) + A-(aq) acid base  H K eq =     O H O H Since initial  2 2  K eq = conjugate acid       HA eq will we ,      c HA conjugate base + − O 3 A eq eq  H O 2 assume   c O = 2 that H + − H O 3 A eq eq eq

  26. In case of weak acid       c HA eq   eq HA     eq HA + − H O 3 A = eq eq K eq    A + − H O 3 eq eq =   = K K c a eq + − H O 3 A = eq eq K a Ionization constant of the acid

  27. In case of weak acid       eq HA + − H O 3 A eq eq = K a [H3O+] =[A-]  O H =  O H +2 3   eq HA  = +   + O e H  eq q 3 3  K a   K acid a [acid k a ] [H3O+] =[H+] = pH= - log [H+ ]

  28. Example Calculate the pH of 0.1 M CH3COOH. The dissociation constant of acetic acid is 1.8 × 10– 5.

  29. In case of weak base [base K b ] [OH-]= pOH = -log [OH-] pH = 14 – pOH

  30. Example Calculate the pH of 0.1 M NH3solution. The ionization constant, Kb, for NH3is 1.8 × 10– 5.

  31. pH of salts Strong acid and weak base, e.g, NH4Cl Weak base and weak acid, e.g, CH3COONH4 From strong base and strong acid, e.g, NaCl Weak acid and strong base,e.g, CH3COONa [H+] = K • WKK [H+] = a b [H+] = pH= 7 pH= -log [H+] pH= -log [H+] pH= -log [H+]

  32. In case of salts 1) In case of salt derived from: Strong acid + strong base salt It is a neutral salt pH = 7 eg. HCl + NaOH NaCl

  33. In case of salts 2) In case of salt derived from: strong base + weak acid salt So Hydrolysis of salt occur + + + Na Ac – Na+ OH- e.g: H.OH AcH Kw . Ka [H+ ] = Csalt Csalt= conc. of salt

  34. Example Calculate the pH of 0.625 M solution of CH3COONa. Ka 1.754 × 10– 5

  35. In case of salts 3) In case of salt derived from: weak base + strong acid salt NH4Cl + H.OH H ++ Cl -+ NH4OH e.g (hydrolysis ) + [H+]= pH = - log [ H ]

  36. Example Calculate the pH of a 0.20 M solution of ammonium chloride, Kb = 1.8 × 10– 5.

  37. In case of salts 4) In case of salt derived from : weak acid + weak base salt The pH depends only on value of Ka& Kb K • WKK a ] [ H+ pH= - log ] [ = H+ b

  38. Example Calculate the pH of a solution of ammonium acetate. Given that : Ka = 1.75 × 10–5, Kb = 1.8 × 10–5and Kw = 1.0 × 10–14

  39. By the end of this lecture you will be able to ➢ Define a self ionization of water ➢ Define a pH of a solution ➢ Calculate pH of different solutions • • • • • Strong acid strong base Weak acid Weak base Solution of salt

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