pH and percent dissociation problem?

Question

Acrylic acid (CH2=CHCO2H) is a precursor for many important plastics. Ka for acrylic acid is 5.6 10-5.

(a) Calculate the pH of a 0.40 M solution of acrylic acid.

(b) Calculate the percent dissociation of a 0.40 M solution of acrylic acid.

(c) Calculate the [H+] necessary to ensure that the percent dissociation of a 0.40 M solution of acrylic acid is less than 0.010%.

(d) Calculate the pH of a 0.057 M solution of sodium acrylate (NaC3H3O2).

Answer 1

I see that you have several similar questions.
I'll solve this for you in detail so that you can work on the others yourself.

Like in the other question you have again a weak monoprotic acid. So I'll write it for simplicity HA where A is the CH2=CHCOO group

(a)
.. .. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. .. .. C
Dissociate .. .. x
Produce .. .. .. .. .. .. .. x .. .. x
At equil .. .. C-x .. .. ..x .. .. x

Ka= [H+][A-]/[HA] = x^2/(C-x) with C=0.4 M
Let's assume that 0.4 >> x and 0.4-x=0.4
Then the equation is simplified to
Ka= x^2/0.4 => x= squareroot (0.4*Ka) =>
x= SQRT(0.4*5.6*10^-5) = 4.73*10^-3. It is not too much smaller than 0.4 but it is not close enough either so let's get an even more exact solution by solving the quadratic though it might not be really necessary
x^2=0.4*Ka-Kax =>
x^2 + (5.6*10^-5)x- 2.24*10^-5 =0
x1= 4.70*10^-3
x2=-4.76 *10^-3 <0 thus rejected. and the answer is x=4.70*10^-3 (in this case, we could have gotten away with the assumption)

pH=-logx= -log(4.7*10^-5) =4.33

(b) a=x/C where C=0.4 in our case. If they asked you for the value of a for a different C or if you hadn't calculated x from (a), then you would say that x=aC and Ka=(aC)^2/(C-aC)= a^2C/(1-a), solve the quadratic and find a. In this case we know both x and C, so we can easily determine a.
a=x/C =(4.7*10^-5)/0.4 = 0.000117 = 0.0117%

(c) You want maximum a=0.01%= 0.0001 =10^-4
we said x=aC= 10^-4*0.4 =4*10^-5.
Now we added a bit of another acid in order to reduce a, so

.. .. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. .. .. 0.4 .. .. ..y
Dissociate .. .. x
Produce .. .. .. .. .. .. .. x .. .. x
At equil .. .. 0.4-x .. .. y+x .. .x

Ka= x[H+]/(0.4-x)
again x=aC= 4*10^-5 =>
[H+]=(0.4-x)*Ka/x =(0.4-4*10^-5) *(5.6*10^-5) /(4*10^-5)=0.55994 M you need to have at equilibrium. If they asked you how much acid you would have to add then
y=[H+] - x= 0.55994 - 0.00004= 0.5599
Practically, you need to add so much H+ that [H+]=y= 0.56 M

(d) You will have hydrolysis

.. .. .. .. .. .. .. A- + H2O <=> HA +OH-
Initial .. .. .. 0.057
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. x .. .. x
At Equil. .. 0.057-x .. .. .. .. .. .. x .. .. x

Kb= [HA][OH-]/[A-] =x^2/(0.057-x)
Kb=Kw/Ka =(10^-14)/(5.6*10^-5) =1.79*10^-10

Let's do the assumption 0.057 >> x
then x^2/0.057=Kb =>
x =SQRT(0.057*1.79*10^-10)= 3.2*10^-6 so our assumption is valid
pH = 14-pOH=14-(-logx) =14-(-log(3.2*10^-6)) = 8.50