How do buffers maintain pH?

2 Answers
May 27, 2016

Buffers moderate both [H_3O^+] and [HO^-].

Explanation:

The weak acid HA undergoes an acid base equilibrium in water according to the equation:

HA(aq) + H_2O(l) rightleftharpoons H_3O^+ + A^-

As with any equilibrium, we can write the equilibrium expression:

K_a = ([H_3O^+][A^-])/([HA])

This is a mathematical expression, which we can divide, multiply, or otherwise manipulate PROVIDED that we do it to both sides of the expression. Something we can do is to take log_10 of BOTH sides.

log_10K_a=log_10[H_3O^+] + log_10{([A^-]]/[[HA]]}

(Why? Because log_10AB=log_10A +log_10B.)

Rearranging,

-log_10[H_3O^+] - log_10{[[A^-]]/[[HA]]}=-log_10K_a

But BY DEFINITION, -log_10[H_3O^+]=pH
, and -log_10K_a=pK_a.

Thus pH=pK_a+log_10{[[A^-]]/[[HA]]}

Do not be intimidated by the log function. When I write log_ab=c, I ask to what power I raise the base a to get c. Here, a^c=b. And thus log_(10)10=1, , log_(10)100=2, log_(10)10^(-1)=-1 . And log_(10)1=0

Given our equation, it tells us that when [A^-]=[HA], then pH=pK_a. Why? Because log_(10)1=0, because 10^0=1. When acid is added to the solution, pH decreases only moderately, because the acid protonates the A^- already present in solution.

Buffers thus moderate pH, and keep pH close to the pK_a of the weak acid initially used.

I acknowledge that I have hit you with a lot of facts. But back in the day A level students routinely used log tables before the advent of electronic calculators. If you can get your head round the logarithmic function you will get it.

Dec 21, 2017

Buffers react with added acid or base to resist a change in pH.

Explanation:

Buffers are special solutions that react with added acid or base to limit the change in pH levels.

For instance, carbonic acid is a weak acid that does not dissociate completely while in water - a small amount is dissociated into "H"^"+" ions and hydrogen carbonate anions (conjugate base).

"H"_2"CO"_3 โ‡Œ "H"^"+" + "HCO"_3^"-"

If you add "H"^"+" ions to the solution, the conjugate base will react with them - reforming the weak acid - keeping the concentration of "H"^"+" ions (and thus the pH) nearly constant.

If you add "OH"^"-" ions to the solution, they will react with the "H"^"+" ions to form water. However, the position of equilibrium will shift to the right โ€” again keeping the concentration of "H"^"+" ions and the pH nearly constant.