# 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

#"pH" ~~ 2.09#

In principle, it does not matter much which acid you allow to go first, as long as the smaller

(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

Since these are both weak acids that "have" similar

#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

For

#"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

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

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

We do the small

#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"(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

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

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

Here we see that

Therefore, **in the presence of**

#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

#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

That makes the

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