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Weak acids and bases Salt of weak acid and bases buffer Lecture 9 9 Feb 2011. Noorulnajwa Diyana Yaacob noorulnajwa@unimap.edu.my. Weak Acid. A weak acid is an acid that does not completely donate all of its hydrogens when dissolved in water.

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weak acids and bases salt of weak acid and bases buffer lecture 9 9 feb 2011

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

NoorulnajwaDiyanaYaacob

noorulnajwa@unimap.edu.my

weak acid
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.
weak acid1
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:
slide4

Example:

Calculate the pH and pOH of a

1.0 x 10-3 solution of acetic acid

slide5

Solution :

HOAc H+ + OAc-

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

[HOAc]

slide6

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]

weak bases
Weak Bases
  • A weak base is a chemical base that only partially ionize in water.
  • Refer Example 7.8 for more understand
salts of weak acids and bases
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.
salts of weak acids and bases1
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

slide11

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:
slide12

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,
slide13

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
slide14
Example

Calculate the pH of a 0.10 M solution of sodium acetate

solution
Solution:
  • Write the equilibria:
  • Write the equilibrium constant
slide17

Similar equations can be derived for the cations of salts of weak base.

  • Refer Example 7.10 for more understanding
what is buffers
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.
slide20

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.
buffer solutions
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

slide22

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 :
slide23

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

  • Inverting the last term, it becomes (+ve)
slide24

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
slide25

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.

slide26

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

slide27

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

buffering mechanism
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.

slide30

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.
slide31

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⁻.
slide32

↑ 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
slide33

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 :
slide34

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
slide35

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.

slide36

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.

slide37

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.

slide38

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:
slide40

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

slide41

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