Question

Find the change in `"pH"` that occurs when `"0.010 mol NaOH"` is added to `"1.0 L"` of a buffered solution containing `"0.50 M"` acetic acid and `"0.50 M"` sodium acetate?

The `K_a` of acetic acid is `1.8 xx 10^(-5)`.

Answer 1

Well...`pH=pK_a + log_10([[""^(-)OAc]]/[[HOAc]])`

Explanation:

And so initially we gots...

`pH=underbrace(-log_10(1.80xx10^-5))_"4.74"=4.74`

Why....because `log_10([[""^(-)OAc]]/[[HOAc]])=log_10(1)=0`

But when we add the hydroxide to the buffer, some of the acetic acid is deprotonated, and clearly, the `[""^(-)OAc]` is slightly increased...

And so, after the `0.010*mol` `NaOH` is added, we gots...

`log_10([[0.50+0.010]]/[[0.50-0.010]])=log_10(1)=+0.017`

...the new `pH=4.76`...the action of the buffer resists a GROSS change in solution `pH`...`pH` is raised SLIGHTLY upon addition of the base...