Definition of Buffers (acid-base)

Buffer solutions resist pH changes.

A buffer is a solution that can maintain a nearly constant pH if it is diluted, or if relatively small amounts of strong acids or bases are added.

A buffer solution can be made by mixing a weak acid with one of its salts OR mixing a weak base with one of its salts.

A more technical way of saying this is that a buffer solution consists of a mixture of a weak acid and its conjugate base OR a weak base and its conjugate acid.

Consider a solution containing both acetic acid, CH₃COOH, and acetate ions, CH₃COO^-. (These could alternatively be named ethanoic acid and ethanoate ions.) This solution can be made by dissolving sodium acetate (sodium ethanoate) in acetic acid (ethanoic acid).

Any strong base that is added to the solution is neutralized by the acetic acid, which acts to remove OH^-:

CH₃COOH _(aq) + OH^-_(aq) CH₃COO^-_(aq) + H₂O _(aq)

Any strong acid that is added to the solution is neutralized by the acetate, which acts to remove H⁺:

CH₃COO^-_(aq) + H⁺_(aq) CH₃COOH _(aq)

The amount of strong acid or base that a buffer can neutralize is called the buffer capacity.

Since the buffer concentration is usually high, the hydrogen ion concentration [H⁺] can be calculated from the equilibrium expression assuming that the concentration of the conjugate acid-base pair does not change appreciably.

Example
What is the pH of a buffer solution containing 0.100 moles of both CH₃COOH and CH₃COO^- in 0.100 liters of water?

The equilibrium is:

CH₃COOH _(aq) CH₃COO^-_(aq) + H⁺_(aq)        K_a = 1.8 x 10^-5 (at 25°C)

We use the following equation to find pH:

pH = pK_a - p([conjugate base]/[acid])

Remembering that p is -log₁₀, we can write:

pH = -log₁₀K_a + log₁₀([conjugate base]/[acid])

Therefore:

pH = -log₁₀1.8 x 10^-5 + log₁₀(1.00 M/1.00 M)

Therefore:

pH = 4.74

Notice that when the concentrations of the weak acid and conjugate bases are equal, the pH of the buffer solution equals pKa.

Let's say we add 10.0 ml of 1.00 M HCl to the buffer solution.

The CH₃COO^- ions in the buffer solution react with all of the H⁺ from the HCl to form CH₃COOH.

The effect is that the concentration of CH₃COOH increases and the concentration of CH₃COO^- decreases.

The equilibrium is:

CH₃COOH _(aq) CH₃COO^-_(aq) + H⁺_(aq)        K_a = 1.8 x 10^-5 (at 25°C)

So:

            [H+][CH3COO-]
Ka  =  ------------------------
              [CH3COOH]
				 		 
		

To calculate pH we need to find [H⁺], so we'll rearrange the above equation to get:

               Ka [CH3COOH]
[H+]  =  ------------------------
                 [CH3COO-]
			  
		

Since the volume of the weak acid and its conjugate base are always the same, we can modify the equation to read:

             Ka (moles of CH3COOH)
[H+]  =  ------------------------
             (moles of CH3COO-)
			  
		

The changes in moles that take place after the addition of the HCl are summarized as follows:

	CH3COOH(aq)	H+(aq)	CH3COO-(aq)
Initial Amount	0.1 moles	(0.010 litres)(1.00 M) = 0.010 moles	0.1 moles
Change in Amount	+ 0.01 moles	-0.01 moles	- 0.01 moles
Equilibrium Amount	0.101 moles	?	0.09 moles

Therefore, we can now put the following values in our equation:

             1.8 x 10-5 x (0.101 moles)
[H+]  =  ------------------------
             (0.09 moles)
						 				 
	[H+] = 2.02 x 10-5
	
We know that pH = -log10[H+]	

Therefore, pH = -log10(2.02 x 10-5) = 4.69

Adding a small amount of strong acid to the buffer solution has barely changed the solution's pH from its original value of 4.74.