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?

1 Answer
anor277
Mar 16, 2017

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.

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