Question

Using acetic acid and sodium acetate solution, describe three different methods to produce a buffer with equal ability to resist strong acid/strong base addition.

Answer 1

The different methods which can be adopted to prepare acetic acid - acetate buffer solution with same capability to resist pH changes when strong acid / base is added differ in adjustments of acid & salt concentrations as the pH of a buffer solution is given by Hasselbalch- Handerson's equation,

..................pH = pK_a   + log { [ salt ] /.[acid ] ,

1.

Direct method - In this method the concentrations of acid and salt required for preparing a buffer solution of desired pH are calculated.

Steps

(i) here we first find [ H⁺ ] corresponding to the required pH using relation - , pH = -log [ H⁺]

therefore , [H⁺ ] = antilog ( -pH )

(ii) Then we apply the equilibrium expression - , K_a   = [H⁺ ] [ A^- ] / [ HA ]

....................................................................[ H⁺  ] = K_a x [HA] / [ A^- ]

Knowing the values of [ H⁺ ] , [HA] , and K_a we can calculate the ratio [ HA] / [ A^- ] or [ A^- ] from the above expression.

(iii) Using the above calculations / informations we can adjust the concentrations of given salt( CH3COONa) solution or acid (CH3COOH ) solution to prepare desired volume of buffer solution with definite pH

2

Alternatively ,

(i) we can prepare a solution of pH = 4.7445 using equimolar concentrations of salt aswell as acid

since according to Handerson's equation -

p H = pK_a   + log { [ Salt ] / [ acid ] } , and when [ salt ] = [ acid ]

pH = p K_a   = 4.7447 .......................pK_a for CH3COOH is known to be = 4.7447

(ii) Using this solution we can prepare a buffer solution of desired pH by any of the following methods-

(iii) Addition of strong acid. (say HCl ) of known concentration-

The prepared buffer solution of pH has certain concentration ( say a moles per litre ) of weak acid HA and b moles per litre of the salt Na⁺ A^- ie.

..........................HA <========> H⁺   + A^-

............................a....................................b......

When x moles per litre of HCl is added, it forms x moles per litre of H⁺ ions. These combine with H⁺ions already present in the solution , and equilibrium is pushed to the left . Thus nowthe concentration of A^- is decreased by x moles / L and that of acid HA is increased by x moles per litre. So,

............................HA <========> H⁺   + A^-  

...........................( a +x )..........................(b - x)..........

Thus we can write equilibrium expression for buffer solution with changed concentrations as

..................................K_a   = [ H⁺ ] [ A^- ] / [ HA ]

...........................................= [ H⁺ ] ( b - x ) / ( a + x)

...............therefore [ H⁺ ] = K_a ( a + x ) / ( b - x )

Knowing the value of K_a   , [ H⁺ ] of the buffer can be calculated ,and then

(iv ) the concentrations of acid and salt is adjusted so as to get a buffer solution of desired pH.

3.

By addition of a strong base of known concentration to the buffer solution already prepared

(i) When NaOH solution is added to the buffer solution the equilibrium shifts to the right till x moles per litre of H⁺ are produced that are needed to netralize the added OH^- .ions.The net result in this case can be expressed as

...................HA <===========> ....H⁺   + A^-   

...................( a-x )..................................( b+ x )

hence ,       ................K_a   = [ H⁺ ] ( b + x ) / ( a - x )

and ............................[ H⁺ ] = K_a ( a - x ) / ( b + x )

(ii) now finally knowing the concentration of H⁺ and K_a   under changed conditions (ie. after adding NaOH solution,

the volume of NaOH solution needed to get a buffer with desired pH can be calculated.