HYDROLYSIS

By:Shibghatullah Muhammady

XI Exact 1/30/13625

What is hydrolysis?

Hydrolysis is analyzing of salt by water becomes its acid and base.

It’s the opponent reaction of salting reaction.

Base + AcidSalting

HydrolysisSalt + Water

We have four kinds of salt solution based on the origin.1. Salt solution that formed from strong acid and

strong base.2. Salt solution that formed from strong acid and

weak base.3. Salt solution that formed from weak acid and

strong base.4. Salt solution that formed from weak acid and weak

base.

How is hydrolysis done?

A. Salt solution that formed from strong acid and strong base.

•There won’t be hydrolysis. •Example :I. NaCl Na+ + Cl-

II. 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.

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.

B. Salt solution that formed from strong acid and weak base.B. Salt solution that formed from strong acid and weak base.

There’ll be hydrolysis because positive ion There’ll be hydrolysis because positive ion of base will bound with ion OHof base will bound with ion OH-- from water from water and it forms base.and it forms base.

[H[H++] > [OH] > [OH--] ] pH < 7 pH < 7

Example : NaCl solutionExample : NaCl solution

I. NH4Cl NH4+ + Cl- (α NH4Cl =1)

II. H2O OH- + H+ (α H2O <1)From these equilibrium reaction, there are two equilibrium reaction.

III. NH4+ + OH- NH4OH (α NH4OH <1)

IV. 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+

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])

The way to look for pH

C. Salt solution that formed from weak acid and strong base.C. Salt solution that formed from weak acid and strong base.

There’ll be hydrolysis because There’ll be hydrolysis because negative ion of acid will bound with negative ion of acid will bound with ion Hion H++ from water and it forms acid. from water and it forms acid.

[H[H++] < [OH] < [OH--] ] pH > 7 pH > 7

Example : CHExample : CH33COONaCOONa

I. CH3COONa Na+ + CH3COO- (α CH3COONa = 1)

II. H2O OH- + H+ (α H2O <1)From these equilibrium reaction, there are two equilibrium reaction.

III. Na+ + OH- NaOH (α NaOH =1)

IV. H+ + CH3COO- CH3COOH (α CH3COOH <1)The most influence is done by reaction II and IV.

II. H2O OH- + H+

IV. H+ + CH3COO- CH3COOH

V. CH3COO- + H2O CH3COOH + OH-

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])

The way to look for pH

D. Salt solution that formed from weak acid and weak base.D. Salt solution that formed from weak acid and weak base.

There’ll be total hydrolysis because positive There’ll be total hydrolysis because positive and negative ions will bound with OHand negative ions will bound with OH-- and H and H++ from water and it forms their acid and base.from water and it forms their acid and base.

The amount of bounded both ions (OHThe amount of bounded both ions (OH-- and H and H++) ) is influenced by the power of acid and base is influenced by the power of acid and base that form salt.that form salt.

KKaa ~ (1/[~ (1/[HH++] ] boundedbounded) ~ ) ~ α α ~ [~ [HH++] ] freefree ~ (1/ ~ (1/pHpH) and ) and KKbb ~ (1/[~ (1/[OHOH-- ] ] boundedbounded) ~ ) ~ α α ~ [~ [OHOH--] ] freefree ~ (1/ ~ (1/pOHpOH))

Example : NHExample : NH44CN solutionCN solution

I. NH4CN NH4+ + CN- (α NH4CN = 1)

II. H2O OH- + H+ (α H2O <1)From these equilibrium reaction, there are two equilibrium reaction.

III. NH4 + + OH- NH4OH (α NH4OH <1)IV. 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

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)

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)

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))

• 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.

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