Question

What is the `pH` of a buffer solution prepared from `50*mL` of sodium hydroxide solution that is `0.220*mol*L^-1` with respect to `NaOH`, and `100*mL` of acetic acid solution that is `0.150*mol*L^-1` with respect to `HOAc`?

Answer 1

`pH=5.20`

Explanation:

We use the buffer equation such that,

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

But of course, we have to calculate `[""^(-)OAc]` and `[HOAc]`. The volume of the solution is clearly `150*mL`.

`"Moles of NaOH"-=(50.0xx10^-3Lxx0.220*mol*L^-1)=0.011*mol.`

`"Moles of HOAc"-=(100.0xx10^-3Lxx0.150*mol*L^-1)=0.015*mol.`

And since `NaOH` reacts quantitatively with `HOAc`, we have a solution that is nominally,

`[HOAc]=((0.015-0.011)*mol)/(150xx10^-3L)=0.0267*mol*L^-1`

And `[""^(-)OAc]=(0.011*mol)/(150xx10^-3L)=0.0733*mol*L^-1`

And so we fill in the dots; the `pH` should be elevated from the `pK_a` (which I happen to know is `4.76`; these data really should have been supplied with the question!).

`pH=4.76+log_10{(0.0733*mol*L^-1)/(0.0267*mol*L^-1)}`

`=4.76+0.439=5.20`.