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?

1 Answer
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 #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)#