BUFFER SOLUTION

Solution SBI 2008 42

Buffer solution is a solution which is composed from weak acid and its salt or weak base and its salt.

Example:1. CH3COOH and CH3COONa

This buffer solution composed of weak acid (CH3COOH) and its salt (CH3COONa)

2. NH4OH and NH4Cl

This buffer solution composed of weak base (NH4OH) and its salt (NH4Cl)

BUFFER SOLUTION(cont’d)

Solution SBI 2008 43

In the first solution, if we add acid, then the acid will be neutralized by left over of acid (sisa asam)

CH3COO- + H+ CH3COOH

If we add base (OH-), it will be neutralized by the acid; CH3COOH + OH- CH3COO- + H2O

The same thing occur when we add acid or base to the buffer solution 2 ( NH4OH and NH4Cl)

To calculate pH of the two buffer solutions the formula used as the following:

BUFFER SOLUTION(cont’d)

Weak acid and its salt

[H+] = Ka .

[Acid]

[Salt]

- log [H+] = - log Ka - log

[Acid]

[Salt]

So pH will be equaled to:

pH = pKa + log [Acid]

[Salt]

Solution SBI 2008 44

Weak base and its salt

[OH-] = Kb .

[Base]

[Salt]

- log [OH-] = - log Kb - log

[Base]

[Salt]

So pOH will be equaled to:

pOH = pKb + log [Salt]

[Base]

and remember pH + pOH = pKw

BUFFER SOLUTION(cont’d)

Solution SBI 2008 45

[Acid] = the molar concentration of weak acid[Base] = the molar concentration of weak base[Salt] = the molar concentration of salt either

from weak acid or weak base

Example:1. 0.1 M acetic acid mixed with 0.01 M sodium

acetate. If Ka is 1 x 10-5, the pH of this solution will be:

BUFFER SOLUTION(cont’d)

Solution SBI 2008 46

Answer:This solution is composed of weak acid and its

salt, so the buffer formula applies to calculate pH of this solution.

[Acid] = 0.1 M; [Salt] = 0.01 MThe pH = pKa + log [Salt]/[Acid]

pH = - log 1 x 10-5 + log (0.01/0.1) = 5 – 1 = 4

BUFFER SOLUTION(cont’d)

Solution SBI 2008 47

2. 0.01 M ammonium hydroxide mixed with 0.1 M ammonium chloride. If Kb is 1 x 10-5, the pH of this solution will be:

Answer:This solution is composed of weak base

(NH4OH) and its salt (NH4Cl), so the buffer formula applies to calculate pH of this solution.

[Base] = 0.01 M; [Salt] = 0.1 MThe pOH = pKb + log [Salt]/[Base]

BUFFER SOLUTION(cont’d)

Solution SBI 2008 48

The pOH = pKb + log [Salt]/[Base] pOH = - log 1 x 10-5 + log (0.1/0.01)

= 5 +1 = 6

pH = pKw – pOH = 14 – 6 = 8

Question 1. 6 grams of acetic acid (Mr 60) and 9.8 grams

of potassium acetate (Mr 98) is mixed in 1 L solution, what the pH of the solution will be? (Ka 1 x 10-5)

BUFFER SOLUTION(cont’d)

Solution SBI 2008 49

Question 2. 35 grams of NH4OH (Mr 35) and 26.75 grams

of NH4Cl (Mr 53.5) is mixed in 0.5 L solution, what the pH of the solution will be? (Kb 1 x 10-5)

Question 3. If we react 12 g of CH3COOH (Mr 60) and 4 g

NaOH (Mr 40) in 1 L solution, what the pH of the solution (Ka 1 x 10-5)

BUFFER SOLUTION(cont’d)

Solution SBI 2008 50

Question 4. One RSBI student wants to make a buffer

solution of NH4OH (Mr 35) and NH4Cl (Mr 53.5) with the pH of 10 in 500 mL. If Kb is 1 x 10-5, what the mass ratio of base and salt must be added?

HYDROLYSIS

Solution SBI 2008 51

Hydrolysis is a reaction between a salt and water in which the anion (from the weak acid) or the cation (the weak base) of the salt react with water so the solution will be acidic or basic

The hydrolysis can only occur when: the number of moles reacts between acid and base are equal

The solution become basic or acidic depends on the salt that was hydrolyzed.

Hydrolysis(Cont’d)

Solution SBI 2008 52

1. The reaction between salt formed of strong acid and strong base with water

Example as in:NaCl Na+ + Cl-

Na+ is from NaOH Cl- is from HCl

As Na+ and Cl- are from strong base and acid, so NaCl is neutral, then the Na+ and Cl- ions will not be able to react with water.

Hydrolysis(Cont’d)

Solution SBI 2008 53

As a result, the hydrolysis in not occurred to the salt which are formed strong acid and strong base.

How about KCl or MgSO4?

Are they hydrolyzed?

They can not be hydrolyzed as well, as they are formed from strong acid and base.

Hydrolysis(Cont’d)

Solution SBI 2008 54

2. The reaction between salt formed of strong acid and weak base with water

Example as in: NH4Cl

In water, NH4Cl will be ionized as follows:

NH4Cl <==> NH4+ + Cl- (1)

NH4+ + H2O <==> NH4OH + H+ (2)

The NH4OH formed is a based, so the equilibrium constant of the last reaction (in equation 2) will be

Hydrolysis(Cont’d)

K = [H2O]

K = [NH4OH

+] [H

+]

[NH4+] [H2O]

[NH4OH+] [H

+]

[NH4+]

Solution SBI 2008 55

This constant value is given a symbol of Kh. This value is dependent on the salt and temperature.

The value of K. H2O is also a

constant value, as the concen-

tration of H2O as the solvent

is constant.

Kh = [NH4OH

+] [H

+]

[NH4+]

(3)

(4)

(5)

Hydrolysis(Cont’d)

Solution SBI 2008 56

As NH4OH is a weak base which is difficult to ionize, the equilibrium in Equation 2 will be shifted to the right, so the water ionization will increase, so the amount of H+ formed will increase too, so the solution will be acidic.

The Cl- ion will remain constant, as it will not react with water. Why?

Is there any relation between Kb, Kh and Kw?

In the last equation (5) let’s say, we multiply with [OH-]/ [OH-], so we have

Hydrolysis(Cont’d)

Solution SBI 2008 57

Kh = Kb

Kwor

Kh = [NH4OH]

[NH4+] [OH

-]

[H+][OH

-]

Remember: Kb = [NH4OH]

[NH4+] [OH

-]

and Kw = [H+] [OH

-] ; so

Kh = 1Kb

Kw

Hydrolysis(Cont’d)

Solution SBI 2008 58

If we insert Kh = Kw/Kb into Equation (5) then we get:

The concentration of NH4OH and H+ based on the reaction 2 is equal. By assuming that the amount of NH4

+ which reacted with water is very small compared to the initial concentration of NH4

+ (from NH4Cl), so Equation 6 can be changed into following equation to calculate pH of hydrolysis:

(6) [NH4

+]

[NH4OH] [H+]

Kb

Kw=

Hydrolysis(Cont’d)

Solution SBI 2008 59

If the concentration of [NH4+] = [Salt] in molar

concentration then:

[NH4+] Kb

Kw=

2 [H

+]

[H+] =

2 Kw

Kb. [NH4

+] ; So

Kw

Kb. [NH4

+][H

+] =

Hydrolysis(Cont’d)

Solution SBI 2008 60

Kw

Kb. [Salt][H

+] =

and

pH = 1/2 pKw - 1/2 pKb - 1/2 log [salt]

In this reaction the pH will be acidic. Why?

3.The reaction between salt formed of strong base and weak acid with water

For example is CH3COONa, a salt formed from NaOH and CH3COOH

Hydrolysis(Cont’d)

Solution SBI 2008 61

The process like in the salt of strong acid and weak base occurred.

The solution will be…….The value of Kh = Kw/Ka, so:

Kw

Ka. [Salt][OH

-] =

and

pOH = 1/2 pKw - 1/2 pKa - 1/2 log [salt]

pH = 1/2 pKw + 1/2 pKa + 1/2 log [salt]

or if we calculate pH straight way, we can use:

Hydrolysis(Cont’d)

Solution SBI 2008 62

In the hydrolysis no. 2 and 3, we call as partial hydrolysis as only one ion is hydrolyzed.

4.The reaction between salt formed of weak base and weak acid with water

In this hydrolysis as the salt are from an anion of weak acid and a cation of weak base, so all the anions will react with water. This is called as total hydrolysis.

The value of Kh = Kw/(Ka x Kb), so the value of pH will only depend on the value of Ka and Kb

Hydrolysis(Cont’d)

Solution SBI 2008 63

Example 1.Calculate pH of NH4Cl solution 0.1M (Kb NH4OH 1x 10-5)

Answer:

NH4Cl is a salt formed from ………

pH = ½ pKw – ½ pKb – ½ Log [Salt]= (½ x 14) – (½ x 5) – (½x-1) = 7 – 2.5 +0.5 = 4.5

Hydrolysis(Cont’d)

Solution SBI 2008 64

Example 2.Calculate pH of CH3COONa solution 0.1M (Ka

CH3COOH 1x 10-5)

Answer:

CH3COONa is a salt formed from ………

pH = ½ pKw + ½ pKa + ½ Log [Salt]= (½ x 14) + (½ x 5)+ (½x-1) = 7 + 2.5 -0.5 = 9

Hydrolysis(Cont’d)

Solution SBI 2008 64

Question Calculate pH of :5. NaOCl solution 0.5M (Ka HOCl 3x 10-8)6. 26.75 g NH4Cl (Mr 53.5) in 5 L solution (Kb NH4OH 1 x

10-5)7. 1.764 g NaCN (Mr 49) in 1.5 L solution (Ka HCN 5x10-10)8. 1.2 g CH3COOH (Mr 60; Ka 1 x 10-5) is reacted with 0.08

g NaOH (Mr 40)9. 125 mL NH4OH 0.2 M is reacted with 125 mL HCl 0.2 M

Kb NH4OH 1 x 10-5)