What is the [H] necessary to ensure that the % dissociation of a 0.22 M solution of acrylic acid is < 0.01%

Question

pH of a 0.22 M solution acrylic acid = 2.46. % dissociation of a 0.22 M solution of acrylic acid = 1.6%. The pH of a 0.037 M solution of sodium acrylate = 8.41.

Answer 1

Let's write sodium acrylate as NaA and acrylic acid as HA.

Rationale of the solution
NaA is the salt of the weak acid HA with the strong base NaOH. Thus it will hydrolyse and the equilibrium dissociation constant will be Kb=Kw/Ka
So we will use the data from the NaA solution to find the Ka.
Once we have that, we will go to the data for the HA and find what the [H+] for a=0.01%

In detail

.. .. .. .. .. .. ..A- + H2O <=> HA+ OH-
Initial .. .. .. . C
React .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. x .. .. x
At Equil. .. ..C-x .. .. .. .. .. .. .. .x .. .. x

Kb= [HA][OH-]/ [A-] = x^2/(C-x)
but Kb=Kw/Ka

So Kw/Ka = x^2/(C-x) => Ka= Kw*(C-x)/x^2

and
x= [OH-]= Kw/[H+]= Kw/ (10^-pH) =10^-14/10^-8.41 = 10^-5.59

so
Ka =(10^-14)*(0.037-10^-5.59)/ ((10^-5.59)^2) = 5.6*10^-5

.. .. .. .. .. .. HA <=> H+ +A-
Initial .. .. .. C.. .. .. ..y
Dissociate ..x
Produce .. .. .. .. .. .. x .. ..x
At Equil .. .. C-x .. .. x+y ..x

where y is the amount of H+ that we added to reach the desired [H+]. The degree of dissociation is
a= dissociated / initial =x/C => x=aC

Ka= [H+][A-]/[HA] = [H+]x/(C-x)= [H+]aC/(C-aC) =[H+]a/(1-a) =>
[H+]=Ka(1-a)/a = (5.6*10^-5)*(1-0.0001)/0.0001 = 0.56 M is the equilibrium concentration.

Note that the portion of H+ coming from the dissociation is extremely small (x=aC= 10^-4*0.22 =2.2*10^-5) compared to 0.56 so practically you have to add y=0.56 M.