Weak acids and bases salt of weak acid and bases buffer lecture 9 9 feb 2011
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Weak acids and bases Salt of weak acid and bases buffer Lecture 9 9 Feb 2011. Noorulnajwa Diyana Yaacob [email protected] 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

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


[email protected]

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 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:


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


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


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

  • 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 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

  • 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.

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

  • 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:


    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⁻]


    *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

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