Buffers and how they affect pH

Question

What is a buffer?

Answer 1

I'm adding to the answer provided by @Robin D. since the formatting of the ICE tables is a bit misleading and difficult to follow/understand. My answer will not provide specific examples as done in the other answer, but will try to explain how a buffer works.

A buffer is a solution that will resist minor changes in pH when an acid or a base is added to it.

The composition of a buffer will be either a weak acid and the conjugate base of that weak acid, or a weak base and the conjugate acid of that weak base. NOTE: the acid or the base must be WEAK, not strong.

The way a buffer functions is as follows:

when an acid (H⁺) is added, it will react with the base to form the weak acid, and since the acid is weak, it will not ionize appreciably, thus essentially removing the added acid (H⁺) from solution and resisting any large change in pH

when a base (OH^-) is added, it will react with the weak acid to form H₂O and the conjugate base, thus essentially removing the added base (OH^-) from solution and resisting any large change in pH

This can be seen in the following generic example, where the weak acid is HA, and conjugate base in A^-

If acid (H⁺) is added: A^- + H⁺ ==> HA (weak acid, not ionized much, essentially removes added H⁺)

If base (OH^-) is added: HA + OH^- ==> H₂O + A^- (essentially removes OH^- and A^- is a weak base)

Answer 2

I.              Buffers

A.  Definition: a solution characterized by the ability to __resist_ changes in pH when limited amounts of acid or base are added to it.

B.   Examples

1.   Weak acid and its conjugate _base____. MEMORIZE

2.   Weak base and its conjugate __acid_____. MEMORIZE

3.   A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution.

	HC2H3O2(aq)   ⇄	H+ (aq) +	C2H3O2_ (aq)
Initial		.50	0	.50
Change		-     X	+ X	+ X
Equilibrium		.50 -X	X	.50 + X

Ka =	[ H +1 ][C2H3O2 -1] =	(X)( 0.50 + X) ~	( X) ( 0.50 ) =	1.7 X 10-5
	[ HC2H3O2]	( 0.50 – X )	( .50 )

X = 1.7 X 10^-5 = [H⁺¹]  so pH = - log ( 1.7 x 10^-5)  pH = 4.76955

4.   Calculate the change in pH when 0.010 mole of solid NaOH is added to 1.0 L of the buffered solution in question #3. compare this pH change with the change that occurs when 0.010 mole of solid NaOH is added to 1.0 L of water.

Strategy: acetic acid is a weak acid & sodium hydroxide is a strong base. Sodium is a spectator ion (ignore it). Write out a table showing the acid base reaction:

	HC2H3O2 (aq) +	OH-1(aq)  à	C2H3O2-1 (aq) +	H2O(l)
Initial (M)		.50	.010	.50	Ignore this column
Change (M)		-     .010	-     .010	+ .010
Final (M)		.49	0	.51

This is a limiting reagent table above: ⇑, now, you can make your ICE table

	HC2H3O2 (aq) ⇄	H+ (aq) +	C2H3O2-1 (aq)
Initial		.49	0	.51
Change		-     X	+ X	+ X
Equilibrium		.49 – X	X	.51 + X

Ka =	[ H +1 ][C2H3O2 -1] =	(X)( 0.51 + X) ~	( X) ( 0.51 ) =	1.7 X 10-5
	[ HC2H3O2]	( 0.49 – X )	( .49 )

X = 1.63 X 10^-5 = [H⁺¹] pH = 4.79

If you add the same amount of NaOH to water:

.010 mol NaOH =	.010 M NaOH	pOH = 2.00
1 liter

Then solve for pH: pH + pOH = 14           so pH + 2.00 = 14 so pH = 12.00