Calculate the ratio of acetate to acetic acid for an acetic acid/acetate buffer to which strong base was added to result in a pH of #5.76#? #K_a = 1.8 xx 10^(-5)#.

1 Answer
Truong-Son N.
Jan 2, 2018

Well, the weak acid and conjugate weak base form a buffer...

#"HOAc"# #-># weak acid
#"OAc"^(-)# #-># conjugate weak base

And for buffers, the Henderson-Hasselbalch equation applies.

#"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])#

The ratio of acetate to acetic acid is given by #\frac(["OAc"^(-)])(["HOAc"])#.

#"pH" - "pK"_a = log\frac(["OAc"^(-)])(["HOAc"])#

Thus, with the #K_a# of acetic acid as #1.8 xx 10^(-5)#, the #"pK"_a# is #-log(K_a) = 4.76#, and:

#color(blue)(10^("pH" - "pK"_a) = \frac(["OAc"^(-)])(["HOAc"]))#

#= 10^(5.76 - 4.76)#

#= color(blue)(10.0)#

So there is about a #10#-fold excess of acetate compared to acetic acid. That should make sense, since the #"pH"# is higher than the #"pK"_a#, i.e. the solution is basic so that the conjugate weak base species dominates.

From this we can say that for every #Delta"pH" = pm1#, the concentration of one of the species in the buffer multiplies by #bb10#.

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