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Weak acids and bases Salt of weak acid and bases buffer Lecture 9 9 Feb 2011

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Weak acids and basesSalt of weak acid and basesbufferLecture 99 Feb 2011

NoorulnajwaDiyanaYaacob

- 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.

- 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:

Example:

Calculate the pH and pOH of a

1.0 x 10-3 solution of acetic acid

Solution :

HOAc H+ + OAc-

Ka= [H+][OAc-] = 1.75 x 10-5

[HOAc]

x2 = 1.75 x 10-5

1.0 x 10-3 - x

x is smaller than C, neglect it, therefore,

x2 = 1.75 x 10-5 = 1.32 x 10-4M = [H+ ]

1.0 x 10-3

pH = -log 1.32 x 10-4 = 3.88

pOH = 14.00 – 3.88 = 10.12

Ka =[H+][OAc-] = 1.75 x 10-5

[HOAc]

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

- 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.

- 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

- 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:

- 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,

- 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

- Example
Calculate the pH of a 0.10 M solution of sodium acetate

- Write the equilibria:
- Write the equilibrium constant

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

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

- 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.

- 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.

- 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.

- a solution containing a weak acid and its conjugate base, or

weak acid

conjugate base

weak base

conjugate acid

- 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 :

- Taking the (-ve) logarithm of each of this equation :

- Inverting the last term, it becomes (+ve)

- 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

- 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.

- 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 = 0.033 mmol/mL

For OAcˉ ,

0.10 mmol/mL X 20 mL = MOAcˉ x 30 mL

MOAcˉ = 0.067 mmol/mL

= 4.76 + log 2.0

= 5.06

- Refer to example 7.12 for more understanding

- 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.

- 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.

- 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⁻.

- ↑ concentrations ,↑ acid/base can tolerate
- Buffer capacity of a solution is defined as
β = dCBOH/ dpH

= - dCHA / dpH

- dCBOH and dCHArepresents the number of moles per liter of strong base or acid.
- For weak acid or conjugate base buffer solution of greater than 0.001 M the buffer capacity is approximate by

- 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 0.10 mol/L sodium acetate, the buffer capacity is :

- If we add solid sodium hydroxide until it becomes 0.0050 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

- 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.

Solution :

mmol HOAc = 2.0 + 0.1 = 2.1 mmol

mmol OAcˉ = 2.0 – 0.1 = 1.9 mmol

The change in pH is -0.05.

- 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.

- 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:

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

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 10.00 if the final concentration of

salt is said to be 0.200 M

We want 100 mL 0f 0.200 M NH4Cl.

Therefore :

mmol NH4Cl = 0.200 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

The molarity of concentrated ammonia is 14.8 M

- Refer example 7.15 for ,ore understanding