Question

What is the `"pH"` for `"0.11 M"` `"HF"` dissociating in water at `25^@ "C"`? `K_a = 7.2 xx 10^(-4)`

Is there enough information to do this?

Answer 1

I got `2.07`.

---

Well, we have enough information, so yes.

The main things you'll need to know how to do:

1. Write a weak acid dissociation reaction of `"HF"` in water.
2. Write an ICE table underneath it and fill it out.
3. Write the mass action expression for `K_a`, the acid dissociation constant.
4. Solve for `x`, which tends to require the quadratic formula. In this case, `x` represents `["H"_3"O"^(+)]`.
5. Use `["H"_3"O"^(+)]` to get `"pH"`.

`1)`

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

Since `"HF"` is an acid, it donates an `"H"^(+)` to water, and becomes its conjugate base, `"F"^(-)`.

`2)`

ICE Tables are a way to organize the `bb"I""nitial"`, `bb"C""hange"`, and `bb"E""quilibrium"` concentrations for the reactants and products.

When you are given a concentration of a weak acid or base with no other information, it tends to be the initial concentration.

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

`"I"" ""0.11 M"" "" "" "" "-" "" "" "" "" ""0 M"" "" "" ""0 M"`

`"C"" "-x" "" "" "" "" "-" "" "" "" "+x" "" "" "+x`

`"E"" "(0.11 - x)"M"color(white)(.....)-" "" "" "" "x" M"" "" "" "x" M"`

If any stoichiometric coefficients were greater than `1`, you would see that coefficient in front of the `x` for that component in the reaction.

• Reactants have negative changes in concentration towards equilibrium (because they get consumed!).
• Products have positive changes in concentration (they get... produced!).

So, if you saw `2"HF"` for a reactant, you would put `-2x` for the change. If you saw `3"OH"^(-)` for a product, you would put `+3x` for the change.

`3)`

The mass action expression is just the equilibrium constant as a function of the equilibrium concentrations, with products over reactants, raised to their respective stoichiometric coefficients:

`K_a = (["H"_3"O"^(+)]["F"^(-)])/(["HF"])`

`= (xcdotx)/(0.11 - x) = x^2/(0.11 - x)`

`4)` It's up to you how you want to use `K_a = 7.2 xx 10^(-4)` to solve this:

• Solve for the quadratic equation form, `ax^2 + bx + c = 0` and use the quadratic formula to solve this in full.
• Use the small `bbx` approximation to cross out the `x` for the weak acid reactant (or base, if it's a weak base equilibrium). This works when `K` is small enough, and if it's `10^(-5)` or less, chances are that it'll work well.
• Use the "improved" small `bbx` approximation. This is what I'll do here, and I go into it in more detail here:

`K_a ~~ x^2/0.11`

`=> x^"*" ~~ sqrt(K_a(0.11 - x)) -= ["H"_3"O"^(+)]`

Using the third way, what we'll do is set the first `x` to be `0`, and acquire our first `x^"*"`. Our first "guess" is:

`x_1 = sqrt(7.2 xx 10^(-4)(0.11)) = "0.008899 M"`

Now what we do is recycle `x_1` as `x` and get another `x^"*" = x_2`:

`x_2 = sqrt(7.2 xx 10^(-4)(0.11 - 0.008899))`

`=` `"0.008532 M"`

And if we repeat this several more times:

`x_3 = sqrt(7.2 xx 10^(-4)(0.11 - 0.008532))`

`=` `"0.008547 M"`

`x_4 = sqrt(7.2 xx 10^(-4)(0.11 - 0.008547))`

`=` `"0.008547 M"`

So, our answer has converged upon `["H"_3"O"^(+)] = "0.008547 M"`. This works when `(K_a)/(["HA"]) < 1`, and indeed:

`(7.2 xx 10^(-4))/(0.11) "<<" 1`.

`5)`

So, the `"pH"` is:

`color(blue)("pH") = -log["H"_3"O"^(+)]`

`= -log("0.008547 M"//"1 M")`

`= color(blue)(2.07)`