1. Noorulnajwa Diyana Yaacob noorulnajwa@unimap.edu.my
Weak acids and bases Salt of weak acid and bases buffer Lecture 9 9 Feb 2011
Noorulnajwa Diyana Yaacob

2. Weak Acid
A weak acid is an acid that does not completely donate all of its hydrogens when dissolved in water.
These acids have higher pH compared to strong acids, which release all of their hydrogens when dissolved in water.

3. Weak Acid The acidity constant for acetic acid at 25oC is 1.75 x 10-5
When acetic acid ionizes, it dissociates to equal portion of H+ and OAc- by such an amount will always be equal 1.75 x 10-5
If the original concentration of acetic acid is C and the concentration of ionized acetic acid species (H+ and OAc-) is x, then the final concentrationof each species at equilibrium is given by:

4. Example:
Calculate the pH and pOH of a
1.0 x 10-3 solution of acetic acid

5. Solution :
HOAc  H+ + OAc-
Ka= [H+][OAc-] = 1.75 x 10-5
[HOAc]

6. x is smaller than C, neglect it, therefore,
Equation
HOAc H+
OAc-
Initial (M)
Change (M)
Equilibrium (M)
1.0 x 10-3 -x
1.0 x x +x x
x = 1.75 x 10-5
1.0 x x
x is smaller than C, neglect it, therefore,
x = 1.75 x = 1.32 x 10-4M = [H+ ]
1.0 x 10-3
pH = -log 1.32 x 10-4 = 3.88
pOH = – 3.88 = 10.12
Ka =[H+][OAc-] = 1.75 x 10-5
[HOAc]

7. Weak Bases
A weak base is a chemical base that only partially ionize in water.
Refer Example 7.8 for more understand

8. conjugate bases of weak acids e.g.: HCOO-
Common Weak Acids
Acid
Formula
Formic
HCOOH
Acetic
CH3COOH
Trichloroacetic
CCl3COOH
Hydrofluoric HF
Hydrocyanic
HCN
Hydrogen sulfide
H2S
Water
H2O
Conjugate acids of weak bases
NH4+
Common Weak Bases
Base
Formula
ammonia
NH3
trimethyl ammonia
N(CH3)3
pyridine
C5H5N
ammonium hydroxide
NH4OH
water
H2O
HS- ion
HS-
conjugate bases of weak acids
e.g.: HCOO-

9. Salts of Weak Acids and Bases
The salt of a weak acid for example NaOAc is strong electrolyte, like all salt and completely ionizes.
In addition, the anion of the salt of a weak acid is a Brønsted base which will accept protons.
It partially hydrolyzed in water to form hydroxide ion and the corresponding undissociated acid.

10. Salts of Weak Acids and Bases
The ionization constant for sodium acetate is equal to basicity constant of the salt.
If the salt hydrolyzes that salt is consider as a weak base.
The weaker the conjugate acid, the stronger the conjugate base, that is, the more strongly the salt will combine with a proton, as from the water , to shift the ionization to the right.
Hydrolysis constant

11. The value of Kb can be calculated from Ka of acetic acid and Kw, if we multiply both the numerator and denominator by [H+]:
The quantity inside the dashed line in Kw and the remainder is 1/Ka, hence:

12. The product of Ka of any weak acid and Kb of its conjugate base is always equal to Kw:
For any salt of weak acid HA that hydrolyzes in water,

13. The pH of such a salt (Bronstead base) is calculated as the same manner as for any other weak base When the salts hydrolyzes, it forms an equal amount of HA and OH- ,If the [] of A- is CA-, then,
The quantity x can be neglected , which will generally be the case for such weakly ionizes bases

14. Example
Calculate the pH of a 0.10 M solution of sodium acetate

15. Solution:
Write the equilibria:
Write the equilibrium constant

16. Since COAc > Kb , neglect x compared to COAc. Then,

17. Similar equations can be derived for the cations of salts of weak base.
Refer Example 7.10 for more understanding

18. WHAT IS BUFFERS???
Buffer is defined as a solution that resists change in pH when a small amount of an acid or base is added or when the solution is diluted.
To maintain the pH of the reaction at an optimum value.

20. Buffer solution consists mixture of weak acid and its conjugate base or weak base with it conjugate acid at predetermined concentration or ratios.
That is mixture of a weak acid and its salt or a weak base and its salt.

21. Buffer Solutions A buffer solution can be:
a solution containing a weak acid and its conjugate base, or
CH3COOH (aq) CH3COO‒ (aq) + H+ (aq)
It is known as an acidic buffer solution and it maintains a pH value that is less than 7.
a solution containing a weak base and its conjugate acid.
NH3 (aq) + H2O(l) NH4+ (aq) + OH‒ (aq)
It is known as a basic buffer solution and it maintains a pH value that is greater than 7.
weak acid
conjugate base
weak base
conjugate acid

22. Consider an acetic acid-acetate buffer.
The acid equilibrium that governs the system is :
Now, we have added a supply of acetate ions to the system
The hydrogen ion [ ] is NO LONGER EQUAL to the acetate ion [ ]
The hydrogen ion concentration is equal to :

23. Taking the (-ve) logarithm of each of this equation :
Inverting the last term, it becomes (+ve)

24. This form of the ionization constant equation is called the Handerson-Hasselbalch equation
Useful for calculating the pH of weak acid solution containing its salt

25. Example :
Calculate the pH of a buffer prepared by adding
10 mL of 0.10 M acetic acid to 20 mL of 0.10 M
Sodium acetate.

26. Need to calculate the [ ] of the acid and salt in the solution
Need to calculate the [ ] of the acid and salt in the solution. The final volume is 30 mL:
So,
For HOAc,
0.10 mmol/mL X 10 mL = MHOAc X 30 mL
MHOAc = mmol/mL

27. For OAcˉ , 0. 10 mmol/mL X 20 mL = MOAcˉ x 30 mL MOAcˉ = 0
For OAcˉ , 0.10 mmol/mL X 20 mL = MOAcˉ x 30 mL MOAcˉ = mmol/mL = log 2.0 = 5.06

28. Refer to example 7.12 for more understanding

29. BUFFERING MECHANISM
For a mixture of weak acid and its salt, it can be explain as follows.
The pH is governed by the logarithm of the ratio of the salt and acid
pH = constant + log [A⁻]
[HA]
*if solution is diluted the ratio remains constant
So, the pH of the solution does not change.

30. If small amount of strong acid added it will combined with an equal amount of the A⁻ to convert it to HA.
HA H⁺ + A⁻
Le Chatelier’s principle dictates added H⁺ will combined with A⁻ to form HA.
The change in ratio [A⁻]/[HA] is small and hence the change in pH is small.
If acid added in unbuffered solution (NaCl solution) the pH will decreased markedly.

31. If small amount of strong base is added it will combined with part of HA to form an equivalent amount of A⁻. Again, change in ratio is small.
Buffering capacity : amount of acid or base that can be added without causing a large change in pH.
This is determine by the concentrations of HA and A⁻.

32. ↑ concentrations ,↑ acid/base can tolerate
Buffer capacity of a solution is defined as
β = dCBOH / dpH
= - dCHA / dpH
dCBOH and dCHA represents the number of moles per liter of strong base or acid.
For weak acid or conjugate base buffer solution of greater than M the buffer capacity is approximate by

33. The CHA and CAˉ represent the analytical [ ] of the acid and its salt respectively.
If we have a mixture of 0.10 mol/L acetic acid and mol/L sodium acetate, the buffer capacity is :

34. If we add solid sodium hydroxide until it becomes 0
If we add solid sodium hydroxide until it becomes mol/L, the change in pH is :
In addition to [ ], the buffering capacity is governed by the ratio of HA to Aˉ.
It is maximum when the ratio is unity that is the pH = pKa

35. Example
A buffer solution is 0.20 M in acetic acid and
in sodium acetate. Calculate the change in pH
upon adding 1.0 mL of 0.10M hydrochloric acid
to 10 mL of this solution.

36. Solution : mmol HOAc = 2.0 + 0.1 = 2.1 mmol
mmol OAcˉ = 2.0 – 0.1 = 1.9 mmol
The change in pH is

37. A buffer can resist a pH change even when there is added an amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer.

38. Same goes for the weak base and its salt.
Consider the equilibrium between the base B and its conjugate (BrØnsted) acid :
The logarithmic Henderson-Hasselbalch form is derived exactly as above:

39. Since pOH =pKw -pH, we can also write , the above equation form pKw

40. Example : Calculate the volume of concentrated ammonia and the weight of ammonium chloride you would have to take to prepare 100 mL of a buffer at pH if the final concentration of salt is said to be M

41. We want 100 mL 0f 0. 200 M NH4Cl. Therefore : mmol NH4Cl = 0
We want 100 mL 0f M NH4Cl. Therefore : mmol NH4Cl = mmol/mL × 100 mL = 20.0 mmol mg NH4Cl = 20.0 mmol × 53.5 mg/mmol = 1.07 × 10³ mg So, 1.07 g NH4Cl. Calculate [ ] of NH3 by

42. The molarity of concentrated ammonia is 14.8 M

43. Refer example 7.15 for ,ore understanding