Question

What is the effect of buffer concentration on pH?

Answer 1

Well, it depends on how you compose your buffer....

Explanation:

See this old answer.

`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]]}`

When we have EQUAL CONCENTRATIONS of `A^-` and `HA`, clearly `pH=pK_a`. Why? Because `[A^-]=[HA]`, and thus `log_10([[A^-]]/[[HA]])=log_(10)1=0`.

Most of the time, the `pH` remains tolerably close to the `pK_a` of the weak acid used in the buffer.