Calculate the pH of a solution of 0.10 M #HF# that is also 0.15 M #HCN# if #K_a# for #HF# is #1.0xx10^-5# and #K_a# for #HCN# is #1.0xx10^-7#?

1 Answer
Jun 28, 2018

#"pH" ~~ 2.09#

In principle, it does not matter much which acid you allow to go first, as long as the smaller #K_a# is smaller by at least #10^3#. You may want to try this in reverse order, doing #"HCN"# first and #"HF"# second, just to see if you get about the same #"pH"# again.

(You should get #x = 9.64 xx 10^(-6) "M"# for #["H"_3"O"^(+)]# from #"HCN"# going first, and #x ~~ "0.00812 M"# for #"HF"# going second. However, when #"HF"# goes, the small #x# approximation can only be done with #"0.10 M"#, and you can say that #9.64 xx 10^(-6)# #"<<"# #x# instead in this case.)

Also, a common pitfall is to forget about the #"H"_3"O"^(+)# contributed by the acid that goes first. That should be included in the initial concentration of #"H"_3"O"^(+)# for the acid that goes second, because they suppress each other.


Since these are both weak acids that "have" similar #K_a# values, the weaker one (#"HCN"#) is not supposed to be ignored. But these #K_a# values are not correct...

#K_a("HF") = 6.6 xx 10^(-4)#
#K_a("HCN") = 6.2 xx 10^(-10)#

So, we'd "have" to do sequential ICE tables, but we should not have to for this problem. We SHOULD be able to just ignore #"HCN"#...

For #"HF"#:

#"HF"(aq) + "H"_2"O"(l) rightleftharpoons "F"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.10" "" "" "-" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "-" "+x" "" "" "+x#
#"E"" "0.10-x" "-" "" "x" "" "" "" "x#

Therefore, using its #K_a# (which is actually #6.6 xx 10^(-4)#):

#6.6 xx 10^(-4) = (["F"^(-)]["H"_3"O"^(+)])/(["HF"])#

#= x^2/(0.10 - x)#

We do the small #x# approximation because that was the intent of the problem, and:

#6.6 xx 10^(-4) ~~ x^2/0.10#

#=> x = sqrt(0.10 cdot6.6 xx 10^(-4)) = "0.00812 M" = ["H"_3"O"^(+)]_("HF")#

(the true answer was #"0.00780 M"#, #4.15%# error, which is just within range of good enough.)

That concentration becomes the initial in the next ICE table, which we will find does not change much at all.

For #"HCN"# after #"HF"#:

#"HCN"(aq) + "H"_2"O"(l) rightleftharpoons "CN"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.15" "" "" "-" "" "" "0" "" "" "" "0.00812#
#"C"" "-x" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.15-x" "-" "" "" "x" "" "" "" "0.00812+x#

For this #K_a# (which is actually #6.2 xx 10^(-10)#):

#6.2 xx 10^(-10) = (["CN"^(-)]["H"_3"O"^(+)])/(["HCN"])#

#= (x(0.00812+x))/(0.15 - x)#

Here we see that #6.2 xx 10^(-10)# #"<<"# #0.00812#, or #8.12 xx 10^(-3)#, as well as #0.15#.

Therefore, in the presence of #"HF"#:

#6.2 xx 10^(-10) = (x(0.00812))/(0.15)#

#= 0.0541x#

#=> x = 1.15 xx 10^(-8) "M" = ["H"_3"O"^(+)]_("HCN")#

As a result, the total #["H"_3"O"^(+)]# would be:

#color(blue)(["H"_3"O"^(+)]) = ["H"_3"O"^(+)]_("HF") + ["H"_3"O"^(+)]_("HCN")#

#= "0.00812 M" + 1.15 xx 10^(-8) "M" ~~ color(blue)("0.00812 M")#

(Clearly, nothing much changed, because #"HCN"# is a much weaker acid in reality, by a factor of about #1000000#.)

That makes the #bb"pH"#:

#color(blue)("pH" = -log(0.00812) = 2.09)#