# 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?

Is there enough information to do this?

##### 1 Answer

I got

Well, we have enough information, so yes.

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

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

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

Since

**ICE Tables** are a way to organize the

*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

- 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

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)#

- 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_1 = sqrt(7.2 xx 10^(-4)(0.11)) = "0.008899 M"#

Now what we do is recycle

#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

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

So, the

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

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

#= color(blue)(2.07)#