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
Mar 16, 2017

Answer:

#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#.