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Acid Base Hydrolysis

Acid Base Hydrolysis. Worked Examples. pH. Determine the pH of a solution which contains 4 x 10 -5 M (moles/litre) H + The decimal part is always positive Log 4.0 = 0.602 log (4 x 10 -5 ) = -5 + 0.602 = -4.398 pH = -log[H + ] = -(-4.398) = 4.398 Or log 0.00004 = -4.398 –ve log = 4.398.

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Acid Base Hydrolysis

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  1. Acid Base Hydrolysis Worked Examples

  2. pH • Determine the pH of a solution which contains 4 x 10-5 M (moles/litre) H+ • The decimal part is always positive • Log 4.0 = 0.602 • log (4 x 10-5) = -5 + 0.602 = -4.398 • pH = -log[H+] = -(-4.398) = 4.398 • Or log 0.00004 = -4.398 –ve log = 4.398

  3. pH • Find the hydrogen ion concentration corresponding to pH 5.643 • pH = -log[H+] = 5.643 • log[H+] = -5.643 = (0.357) + (-6.000) • Antilog of 0.357 = 2.28 x 10-6 • [H+] = 2.28 x 10-6

  4. Hydrolysis • Salt of a weak acid and strong base • Calculate: • 1) the hydrolysis constant, • 2) the degree of hydrolysis and • 3) the hydrogen concentration for 0.01 M solution of sodium acetate • Na+ + Ac - + H2O ↔ Na+ + HAc + OH-

  5. Hydrolysis • The degree of dissociation is given by • Substituting for Kh and V = 1/c we obtain • Solving this quadratic equation, x = 0.000235 or 0.0235%

  6. Hydrolysis • If the solution was completely hydrolysed, the concentration of acetic acid produced would be 0.01 M. But the degree of hydrolysis is 0.0235%. Therefore the concentration of acetic acid is 2.35 x 10-6 M • This is also equal to the hydroxide ion concentration • [OH-] = 2.35 x 10-6 • M pOH = 5.63 • pH = pKw – pOH i.e. pH = 14 – 5.63 = 8.37 • The pH may also be determined using • pH = ½pKw +½pKa + ½log c • = 7.0 + 2.37 + ½(-2) • pH = 8.37

  7. Hydrolysis • Salt of a strong acid and weak base • Determine the pH of a 0.2 M NH4Cl solution • NH4- + Cl- + H2O ↔ NH4OH + Cl- + H+ • Since [NH4OH] and [H+] are equal

  8. Hydrolysis • Ammonia in water Kb = 1.85 x 10-5 and • pKb = 4.74 • pH = ½pKw -½pKb - ½log c

  9. Hydrolysis • pH = ½(14) -½(4.74) - ½(-0.6989) • pH = 7.0 – 2.37 + 0.3495 • pH = 4.98

  10. Hydrolysis • Salt of a weak acid and weak base • NH4+ + Ac- + H2O ↔ NH4OH + HAc

  11. Hydrolysis • If x is the degree of hydrolysis of 1g mol. of salt dissolved in V litres of solution, then • [MOH] = [HA] = x/V; [M+] = [A-] = (1 – x)/V • Substituting the values in the equation 1 given in the previous slide • The degree of hydrolysis and consequently pH is independent of the concentration of the solution.

  12. Hydrolysis • Remembering the following equations must hold simultaneously • Also, it can be shown that

  13. Hydrolysis • The hydrogen ion concentration of the hydrolysed solution is calculated in the following manner.

  14. Hydrolysis • If the ionisation constant of the acid and base are equal, i.e. Ka = Kb, pH = ½pKw = 7 and the solution is neutral, although the hydrolysis may be considerable. • Ka > Kb, pH < 7 • Kb > Kc, pH > 7 • The pH of a solution of ammonium acetate is given by: • pH = 7.0 + 2.37 – 2.37 = 7.0 and the solution is approximately neutral • On the other hand, for a solution of ammonium formate • pH = 7.0 +1.88 – 2.37 = 6.51 • Formic acid; Ka = 1.77 x 10-4; pKa = 3.75 and the solution reacts slightly acid.

  15. pH Buffers • A solution of 0.0001 M HCl should have a pH of 4, but this solution is extremly sensitive to traces of alkali from the glass container and ammonia in the air. • Likewise a solution of 0.0001 M NaOH should have a pH of 10 but this solution is sensitive to carbon dioxide in the air. • An aqueous solution of KCl pH = 7 • Likewise, an aqueous solution of NH4Ac pH = 7 • The addition of 1 mL of 1M HCl to 1 litre of KCl changes the pH to ≈ 3 • The addition of 1 mL of 1M HCl to 1 litre of NH4Ac does not change the pH much at all.

  16. pH Buffers • This is because the H+ ions added are mopped up by the acetate ions: H+ + Ac-→ HAc(the equilibrium lies very much to the right hand side of this equation). • The ammonium acetate it is behaving as a buffer and resisting changes in pH. A buffer possesses reserved acidity and reserved alkalinity.

  17. pH Buffers • For a buffer formed from a weak acid (HA) and the salt of a weak acid (MA). HA ↔ H+ + A- • Making an approximation of activities≈ concentrations • If the concentration of the acid = ca • The concentration of the salt = cs • Then the concentration of the un-dissociated portion of the acid = ca – [H+] • The solution must be electrically neutral [A-] = cs + [H+]

  18. pH Buffers • The quadratic equation can be solved for [H+] • It can also be simplified by considering that in a mixture of weak acid with its salt, the dissociation constant of the acid is repressed by the common ion effect. The hydrogen ion concentration is negligibly small in comparison with ca and cs • Equation reduces to:

  19. pH Buffers • Or • If the concentrations of an acid and salt are equal (i.e. half neutralised), then pH = pKa • For acetic acid, Ka = 1.82 x 10-5, pKa = 4.74 • At half titre point a 0.1 M HAc solution would have a pH of 4.74. This would also be true for higher and lower concentrations of HAc, e.g 1.0 M and 0.05 M HAc

  20. pH Buffers • If we add a small concentration of H+ ions they would combine with acetate ions to form acetic acid. • H+ + CH3COO-↔ CH3COOH • Similarly a small concentration of OH- added will combine with H+ ions from dissociation of HAc and form water. More acetic will dissociate to replace the H+ ions depleted in this manner. • Example: Calculate the pH of a solution produced by adding 10 mL of 1M HCl to 1 liter of a solution which is 0.1M in acetic acid and 0.1M in sodium acetate. • The pH of the acetic acid – sodium acetate buffer is given by:

  21. pH Buffers • Neglecting the volume change from 1000 to 1010 mL. • The HCl reacts with acetate ion forming pratically undissociated acetic acid. • H+ + CH3COO-↔ CH3COOH • Therefore [CH3COO-] = 0.1- 0.01 = 0.09 • And [CH3COOH] = 0.1 + 0.01 = 0.11 • pH = 4.74 + log 0.09/0.11 = 4.74 – 0.09 = 4.65

  22. pH Buffers • The addition of the strong acid to the buffer change the pH by 4.74 - 4.65 = 0.09 pH • Adding 10 mL of 1M HCl to 1 litre of pure water (pH 7), the pH would have changed from 7 to –log(0.01) = 2, by 5 pH unites. • A solution that contains equal concentrations of acid and salt, or a half neutralized solution, has the maximum buffer capacity. Other mixtures will also have considerable buffer capacity, but the pH will differ slightly from the half-neutralized acid. • E.g. a quarter neutralized solution of acid, [Acid] = 3[Salt] • pH = pKa + log ⅓ • = pKa – 0.477

  23. pH Buffers • E.g. a three quarter neutralized acid solution, 3[Acid] = [Salt] • pH = pKa + log 3 • = pKa + 0.477 • Generally speaking the buffering capacity is maintained for mixtures within the range: • 1 acid : 10 salt and 10 acid : 1 salt • pH = pKa 1 • The concentration of the acid is usually of the order 0.05-0.20 M. • Similar remarks apply to buffers of weak bases (NH3/NH4Cl)

  24. pH Buffers pH of 0.2 M HAc, 0.2 M NaAc Buffer Mixtures 10 mL Vol.

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