Calculate the #"pH"# of a #"0.32 M"# solution of acetic acid (#K_a = 1.8 xx 10^(-5)#) to three sig figs?

1 Answer
Mar 4, 2017

This #K_a# is on the order of #10^(-5)# or less, so to an even better approximation, #x=[H"^(+)]# is indeed small in comparison to #"0.32 M"#.

Thus, we skip over the ICE table and get:

#K_a = 1.8 xx 10^(-5) = (["H"^(+)]["A"^(-)])/(["HA"])#

#= x^2/(0.32 - x) ~~ x^2/0.32#

#-> ["H"^(+)] = sqrt(K_a[HA]) = 0.0024# #"M"#

and the #"pH"# would just be:

#"pH" = -log["H"^(+)] = 2.620#

To three sig figs it would then be:

#color(blue)("pH" = 2.62)#