Buffers

A buffer is a solution that contains substantial amounts of a compound in both its protonated and deprotanted forms.  As such, it is "resistant" to pH change upon the addition of strong acid or strong base.  This is because the protonated form can neutralize any strong base and the deprotonated form can neutralize any strong acid.  When this happens the ratio of protonated to deprotonated changes, but if their initial amounts are significant, then these changes are small.  The total amount of acid or base that can be added to the buffer before neutralizing all of one of the forms of the compound is its buffer capacity.  The pH of a buffer is generally in a range around the pKa of the acidic form of the compound (or you can think of this as the pOH is around the pKb of the basic form).  So if you want to maintain the pH in a specific region, you should use a buffer where the conjugate pair has a pKa near that value.

For example, let's imagine we wanted to make a buffer with a pH around 3.2, you would then like a compound with a pKa near this value.  Benzoic acid has a pKa of 3.18 so this would be a good choice.  Now you need to make a solution that contains both benzoic acid (the protonated form) and the ion benzoate (the deprotonated form).

There are three ways that we can accomplish this.  The first is to mix a solution of benzoic acid with a solution of a salt that contains the benzoate ions (such as sodium benzoate).  If we put in comparable numbers of moles of benzoic acid and the benzoate ion we'll end up with a buffer solution.  For the pH to exactly match the pKa the concentrations of these two need to be equal.  However, the buffer does not need to have exactly equal concentrations of these two.  We can make a slightly more acidic buffer with more acid (benzoic acid). We can make a slight more basic buffer with more base (benzoate).  Alternatively we can make the buffer by starting with a solution that is benzoic acid and partial neutralizing it to generate the benzoate.  So if I have a 1 mole of benzoic acid in solution and I add 0.5 moles of OH-, the resulting solution will have 0.5 moles of benzoic acid and 0.5 moles of the benzoate ion.   The third way is the opposite of this, I can start with a solution that essentially has all of the compound in the deprtonated from of the benzoate ion and I can partially neutralize this with a strong acid.

Note: In neutralizing the solutions to form the buffer, I cannot neutralize all of the acid or all of the base or I won't make a buffer.

Finally, I can make buffers of different capacity by using different amounts of the compounds.  Let's take the example above of a solution with 1 mole of benzoic acid that is neutralized with 0.5 moles of OH-.  The resulting buffer contains 0.5 moles of both the protonated and deprotonated species.  It has a buffer capacity of 0.5 moles for both acid and base.  Conversely, if I started with a solution with 0.1 moles of benzoic acid and neutralized it with 0.05 moles of OH-, the resulting buffer would have 0.05 moles of each benzoic acid and the benzoate ion.  This buffer would had the same pH as the previous example since there are equal concentrations of protonated (acid) and deprotonated (base) forms.  However, the capacity of this buffer is 10 times less, since there are 10 times fewer moles of each.  The addition of 0.05 moles of strong acid to the first buffer would only make a small change as it would neutralize only ten percent of the base.  The addition of 0.05 moles to the second buffers would neutralize all of the base.

buffer concept » play Which of the following will make a buffer (show/hide answer) The answer is 100 mL of 1M NH3and 50 mL of 1M HCl.This will result in a solution with .05 moles of NH3 and 0.05 moles of NH4+.This is a buffer 100 mL of 1M HFand 100 mL of 2M NaOH.Here the OH- will neutralize all of the HF and there will be excess OH-.This is a strong base solution 100 mL of 1M HCland 50 mL of 1M NaOH.This solution will partially neutralize the HCl, but HCl is a strong acid.You can't make a buffer with a strong acid since the conjugate base is infinitely weak. 100 mL of 1M NH3and 50 mL of 1M NaOH.Here you are adding a strong base to a base. This simply makes a strong base solution. 100 mL of 1M NH3and 50 mL of 2M HCl.Here you completely neutralize all of the base (NH3) and end up with a weak acid solution of NH4+

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