Question

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`?

Answer 1

`"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)`