Solution 5 : buffer solution

BUFFER SOLUTION

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) Solution SBI 2008 42

BUFFER SOLUTION(cont’d)

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: Solution SBI 2008 43

BUFFER SOLUTION(cont’d) Weak acid and its salt +

[H ] = Ka .

Weak base and its salt

[Acid]

-

[OH ] = K b .

[Salt]

+

- log [H ] = - log Ka - log

[Acid]

[Base] [Salt]

-

- log [OH ] = - log Kb - log

[Salt] So pH will be equaled to:

pH = pKa + log

[Salt] [Acid]

[Base] [Salt]

So pOH will be equaled to: pOH = pKb + log

[Salt] [Base]

and remember pH + pOH = pKw

Solution SBI 2008 44

BUFFER SOLUTION(cont’d)

[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:

Solution SBI 2008 45

BUFFER SOLUTION(cont’d)

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 =4

Solution SBI 2008 46

BUFFER SOLUTION(cont’d)

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 The pOH = pKb + log [Salt]/[Base] Solution SBI 2008 47

BUFFER SOLUTION(cont’d)

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) Solution SBI 2008 48

BUFFER SOLUTION(cont’d)

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) Solution SBI 2008 49

BUFFER SOLUTION(cont’d)

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?

Solution SBI 2008 50

HYDROLYSIS

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. Solution SBI 2008 51

Hydrolysis(Cont’d)

1. The reaction between salt formed of strong acid and strong base with water Example as in: NaCl  Na+ + ClNa+ 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. Solution SBI 2008 52

Hydrolysis(Cont’d)

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.

Solution SBI 2008 53

Hydrolysis(Cont’d)

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 Solution SBI 2008 54

+

K=

+

[NH4OH ] [H ] + [NH4 ]

(3)

[H2O] +

K [H2O] =

+

[NH4OH ] [H ] + [NH4 ]

Hydrolysis(Cont’d) The value of K. H2O is also a constant value, as the concentration of H2O as the solvent

(4) is constant.

This constant value is + + [NH4OH ] [H ] given a symbol of Kh. Kh = (5) + This value is [NH4 ] dependent on the salt and temperature. Solution SBI 2008 55

Hydrolysis(Cont’d)

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 Solution SBI 2008 56

Hydrolysis(Cont’d)

Kh =

+

[NH4OH] +

-

[H ][OH ] -

[NH4 ] [OH ] +

Remember: Kb = +

-

[NH4 ] [OH ] [NH4OH] -

and Kw = [H ] [OH ] ; so 1 Kh = Kb

Kw

or

Kh =

Kw Kb

Solution SBI 2008 57

Hydrolysis(Cont’d)

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

+

=

[NH4OH] [H ] +

[NH4 ]

(6)

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: Solution SBI 2008 58

Hydrolysis(Cont’d)

Kw Kb

+ 2

=

+ 2

[H ] =

+

[H ] =

[H ]

+ [NH4 ]

Kw Kb

.

+ [NH4 ]

Kw Kb

; So

. [NH4+]

If the concentration of [NH4+] = [Salt] in molar concentration then: Solution SBI 2008 59

Hydrolysis(Cont’d) +

[H ] =

Kw Kb

. [Salt]

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 Solution SBI 2008 60

Hydrolysis(Cont’d)

The process like in the salt of strong acid and weak base occurred. The solution will be……. The value of Kh = Kw/Ka, so: -

[OH ] =

Kw Ka

. [Salt]

and

pOH = 1/2 pKw - 1/2 pKa - 1/2 log [salt] or if we calculate pH straight way, we can use: pH =

1/2

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

Solution SBI 2008 61

Hydrolysis(Cont’d)

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

Solution SBI 2008 62

Hydrolysis(Cont’d)

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 Solution SBI 2008 63

Hydrolysis(Cont’d)

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 Solution SBI 2008 64

Hydrolysis(Cont’d)

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)

Solution SBI 2008 64