Solution 5 : buffer solution

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+  CH

3COOH

If we add base (OH-), it will be neutralized by

the acid; CH₃COOH + OH-  CH

3COO- + 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]

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

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 M The pH = pKa + log [Salt]/[Acid]

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

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 M

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+ + H

2O <==> 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 K_h. 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 =

[NH

OH

] [H

]

[NH

]

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 K_b, K_h and K_w? In the last equation (5) let’s say, we multiply

Hydrolysis(Cont’d) Solution SBI 2008 57

Kh =

Kb

Kw

or

Hydrolysis(Cont’d)

Solution SBI 2008 58

If we insert K_h = K_w/K_b 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 NH

4Cl), so

Equation 6 can be changed into following equation to calculate pH of hydrolysis:

(6)

[NH₄+]

[NH₄OH] [H+] Kb

Hydrolysis(Cont’d)

Solution SBI 2008 59

If the concentration of [NH₄+] = [Salt] in molar concentration then: [NH₄+] Kb Kw ₌ [H+] 2 [H+] = 2 Kw Kb

.

.

Hydrolysis(Cont’d) Solution SBI 2008 60 Kw Kb

.

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 K_h = K_w/K_a, so:

Kw

Ka

.

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

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

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 K_h = K_w/(K_a x K_b), so the value of pH will only depend on the value of K_a and K_b

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)