What is the pH for 0.11 M HF dissociating in water at 25∘C? Ka=7.2×10−4

Is there enough information to do this?

1 Answer
Truong-Son N.
Jun 8, 2017

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 Ka, the acid dissociation constant.
  4. Solve for x, which tends to require the quadratic formula. In this case, x represents [H3O+].
  5. Use [H3O+] to get pH.

1)

HF(aq)+H2O(l)⇌F−(aq)+H3O+(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 Initial, Change, and Equilibrium 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) + H2O(l) ⇌ F−(aq)+H3O+(aq)

I 0.11 M − 0 M 0 M

C −x − +x +x

E (0.11−x)M.....− 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 2HF for a reactant, you would put −2x for the change. If you saw 3OH− 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:

Ka=[H3O+][F−][HF]

=x⋅x0.11−x=x20.11−x

4) It's up to you how you want to use Ka=7.2×10−4 to solve this:

  • Solve for the quadratic equation form, ax2+bx+c=0 and use the quadratic formula to solve this in full.
  • Use the small x 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 x approximation. This is what I'll do here, and I go into it in more detail here:

Ka≈x20.11

⇒x*≈√Ka(0.11−x)≡[H3O+]

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:

x1=√7.2×10−4(0.11)=0.008899 M

Now what we do is recycle x1 as x and get another x*=x2:

x2=√7.2×10−4(0.11−0.008899)

= 0.008532 M

And if we repeat this several more times:

x3=√7.2×10−4(0.11−0.008532)

= 0.008547 M

x4=√7.2×10−4(0.11−0.008547)

= 0.008547 M

So, our answer has converged upon [H3O+]=0.008547 M. This works when Ka[HA]<1, and indeed:

7.2×10−40.11<<1.

5)

So, the pH is:

pH=−log[H3O+]

=−log(0.008547 M/1 M)

=2.07

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