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HYDROLYSIS

HYDROLYSIS. By: Shibghatullah Muhammady XI Exact 1/30/13625. Salting. Hydrolysis. What is hydrolysis?. Hydrolysis is analyzing of salt by water becomes its acid and base. It’s the opponent reaction of salting reaction. Base + Acid. Salt + Water.

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HYDROLYSIS

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  1. HYDROLYSIS By: Shibghatullah Muhammady XI Exact 1/30/13625

  2. Salting Hydrolysis What is hydrolysis? • Hydrolysis is analyzing of salt by water becomes its acid and base. • It’s the opponent reaction of salting reaction. Base + Acid Salt + Water

  3. We have four kinds of salt solution based on the origin. • Salt solution that formed from strong acid and strong base. • Salt solution that formed from strong acid and weak base. • Salt solution that formed from weak acid and strong base. • Salt solution that formed from weak acid and weak base.

  4. A. Salt solution that formed from strong acid and strong base. • There won’t be hydrolysis. • Example : • NaCl Na+ + Cl- • H2O OH- + H+ NaCl is strong electrolyte  perfectly becomes Na+ and Cl-. H2O is ionizated with small ionization degree (α << 1). Na+ and Cl- don’t influence the equilibrium of water.

  5. H2O OH- + H+ K = [OH-][H+] / [H2O] (α H2O <<1 K. [H2O ] = constant) K . [H2O] = [OH-][H+] Kh = [OH-][H+] Kh= Kw (pH = 7), it’s netral.

  6. B. Salt solution that formed from strong acid and weak base. • There’ll be hydrolysis because positive ion of base will bound with ion OH- from water and it forms base. • [H+] > [OH-]  pH < 7 Example : NaCl solution

  7. NH4Cl NH4+ + Cl- (α NH4Cl =1) • H2O OH- + H+ (α H2O <1) From these equilibrium reaction, there are two equilibrium reaction. • NH4+ + OH- NH4OH (α NH4OH <1) • H+ + Cl- HCl (α HCl =1) The most influence is done by reaction II and III. II. H2O OH- + H+ III. NH4+ + OH- NH4OH V. NH4+ + H2O NH4OH + H+

  8. The way to look for pH • K = [NH4OH ][H+] / [NH4+ ][H2O ] K. [H2O ] = [NH4OH ][H+] / [NH4+ ] (α H2O <<1 K. [H2O ] = constant) Kh = [NH4OH ][H+] / [NH4+ ] Kh = [NH4OH ][H+][OH-] / [NH4+ ] [OH-] Kh = Kw / Kb [NH4OH ] = [H+], so that Kw / Kb = [H+]2 / [NH4+ ] [H+] = ((Kw / Kb). [NH4+ ])½ ([NH4+ ] = [NH4Cl] = [salt]) [H+] = ((Kw / Kb). [salt ])½ pH = ½ (pKw – pKb – log [salt])

  9. C. Salt solution that formed from weak acid and strong base. • There’ll be hydrolysis because negative ion of acid will bound with ion H+ from water and it forms acid. • [H+] < [OH-]  pH > 7 Example : CH3COONa

  10. CH3COONa Na+ + CH3COO- (αCH3COONa = 1) • H2O OH- + H+ (α H2O <1) From these equilibrium reaction, there are two equilibrium reaction. • Na+ + OH- NaOH (α NaOH =1) • H+ + CH3COO- CH3COOH (αCH3COOH <1) The most influence is done by reaction II and IV. II. H2O OH- + H+ IV. H+ + CH3COO- CH3COOH V. CH3COO- + H2O CH3COOH + OH-

  11. The way to look for pH • K = [CH3COOH ][OH- ] / [CH3COO-][H2O ] K. [H2O ] = [CH3COOH ][OH- ] / [CH3COO-] (α H2O <<1 K. [H2O ] = constant) Kh = [CH3COOH ][OH- ] / [CH3COO-] Kh = [CH3COOH ][OH- ][H+] / [CH3COO-][H+] Kh = Kw / Ka [CH3COOH] = [OH- ], so that Kw / Ka = [OH- ]2 / [CH3COO-] [OH-] = ((Kw / Ka). [CH3COO-])½ ([CH3COO-] = [CH3COONa ] = [salt]) [OH-] = ((Kw / Ka). [salt ])½ pH = 14 - ½ (pKw – pKa – log [salt])

  12. D. Salt solution that formed from weak acid and weak base. • There’ll be total hydrolysis because positive and negative ions will bound with OH- and H+ from water and it forms their acid and base. • The amount of bounded both ions (OH- and H+) is influenced by the power of acid and base that form salt. • Ka~ (1/[H+] bounded) ~ α ~ [H+] free ~ (1/pH) and Kb~ (1/[OH-] bounded) ~ α ~ [OH-] free ~ (1/pOH) • Example : NH4CN solution

  13. NH4CN NH4+ + CN- (αNH4CN = 1) • H2O OH- + H+ (α H2O <1) From these equilibrium reaction, there are two equilibrium reaction. • NH4+ + OH- NH4OH (α NH4OH <1) • H+ + CN- HCN (α HCN <1) The most influence is done by reaction II, III, and IV. II. H2O OH- + H+ III. NH4+ + OH- NH4OH (α NH4OH <1) IV. H+ + CN- HCN (α HCN <1) V. NH4+ + CN- + H2O NH4OH + HCN

  14. The way to look for pH • K = [NH4OH ][HCN] / [NH4+ ][H2O ][CN-] K. [H2O ] = [NH4OH ][HCN] / [NH4+ ][CN-] Kh = [NH4OH ][HCN] / [NH4+ ][CN-] Kh = ([NH4OH ][HCN] / [NH4+ ][CN-]) . ([H+][OH-] / [H+][OH-]) Kh = ([H+][OH-]).([HCN] / [H+][CN-]) . ([NH4OH ] / [NH4+ ][OH-])  Kh = Kw/(Ka . Kb)

  15. ni NH4CN = a and its ionization degree is α • nion = αa. • nf = a(1- α)  nfNH4+ = nfCN- = a(1 –α) • nfNH4OH = nfHCN = αa So, • Kh = [NH4OH ][HCN] / [NH4+ ][CN-] = (αa . αa) / (a(1 –α) . a(1 –α)) = α2/(1 –α)2 = (α/(1 –α))2 (used) α = √Kh - α√Kh  α(1 + √Kh) = √Kh  α = √Kh / (1 + √Kh)

  16. Example : HCN H+ + CN- (if nacid = nbase) so, Ka = [H+][CN-] / [HCN] [H+] = Ka . [HCN] / [CN-] [H+] = αa [CN-] = a(1 –α) [H+] = Ka . α / (1 –α) • [H+] = Ka . √Kh • [H+] = Ka . √(Kw / (Ka . Kb)) • [H+] = √ (Kw . Ka / Kb) • pH = ½ (pKw – log (Ka / Kb)) and pOH = ½ (pKw – log (Kb / Ka))  pH =14 - ½ (pKw – log (Kb / Ka))

  17. If one of acid or base is left. • If acid is left  [H+] = Ka . [acid] / [salt] • If base is left [OH-] = Kb . [base] / [salt] But, if |Ka - Kb| so large, the result of these formulas will have large deviation. • This case is not learned in Senior High School.

  18. Thank you May it will be useful for us…

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