Question

If the ratio of weak base to conjugate acid is `10:1`, to what extent does the `"pH"` change and does it increase or decrease?

Answer 1

Suppose the buffer started with a `"pH"` of, well, "`"pH"`", with a `"pKa"` of "`"pKa"`", with general concentrations of acid `"HA"` and conjugate base `"A"^(-)` (or if you prefer, the salt `"NaA"`). Then the Henderson-Hasselbalch equation shows:

`"pH" = "pKa" + log\frac(["A"^(-)])(["HA"])`

Now, if we have `10` times the `"A"^(-)` (noting that no identity of anything changes, so the `"pKa"` of the acid stays the same), we simply get a new `"pH"` of:

`color(blue)("pH"') = "pKa" + log\frac(10["A"^(-)])(["HA"])`

And if we recall... `log(ab) = log a + log b`. Thus...

`=> "pKa" + log\frac(["A"^(-)])(["HA"]) + log 10`

`= "pH" + log 10`

`= color(blue)("pH" + 1)`

So, the `"pH"` increases by `1`, and the solution becomes `10` times more basic. That SHOULD make sense... we did, after all, make it so we had `10` times the conjugate base.