Question

How do buffers maintain pH?

Answer 1

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.

Answer 2

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 ⇌ "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 — again keeping the concentration of `"H"^"+"` ions and the pH nearly constant.