With this information, I have to fill out following chart of molarity and caclul
ID: 974296 • Letter: W
Question
With this information, I have to fill out following chart of molarity and caclulated pH. Im not sure how though
mL of .6M Acetic acid
mL of .6M Sodium Acetate
pH
30
0
2.25
30
10
3.86
0
30
8.83
10
30
5.4
20
20
4.6
Molarity acetic acid
molarity acetate ion
pH calculated
pH from table
% error
2.25
3.86
8.83
5.4
4.6
With this information, I have to fill out following chart of molarity and caclulated pH. Im not sure how though
mL of .6M Acetic acid
mL of .6M Sodium Acetate
pH
30
0
2.25
30
10
3.86
0
30
8.83
10
30
5.4
20
20
4.6
Explanation / Answer
For all these problems, we will need to acid dissociation constant of acetic acid, Ka which is 1.8*10-5.
Acetic acid is a weak acid and it dissociates in solution as
CH3COOH (aq) + H2O ---------> CH3COO- (aq) + H3O+ (aq)
Ka = [CH3COO-][H3O+]/[CH3COOH] (water is neglected since, since it is present in large excess).
After partial dissociation, there is still approximately 0.6 M acetic acid and [CH3COO-] = [H3O+] = c (say).
Therefore,
1.8*10-5 = (c).(c)/(0.6)
or, c2 = 1.8*10-5*0.6
or, c = (1.8*10-5*0.6) = 3.286*10-3
The concentration of proton, i.e, [H3O+] = 3.286*10-3 M.
Therefore, pH = -log10[H3O+] = -log10(3.286*10-3) = 2.48
For the second, fourth and fifth entries in the table, we need to calculate the initial and final concentrations of both acetic acid and acetate base.
We realize that the initial concentration of both acetic acid and acetate are 0.6 M.
Also, we shall find out the final molarity (in the solution, not the equilibrium one) by using the relation
V1 (mL) x S1 (M) = V2 (Ml) x S2 (M)
Volume of 0.6 M acetic acid added (mL)
Volume of 0.6 M acetate added (mL)
Total volume of solution (mL)
Concentration of acetic acid in solution (M)
Concentration of acetate in solution (M)
30
10
40
0.45
0.15
10
30
40
0.15
0.45
20
20
40
0.30
0.30
Next we shall have to set up the ICE Chart, since acetic acid is a weak acid and will undergo dissociation.
Second entry
CH3COOH + H2O -------------> CH3COO- + H3O+
initial 0.45 0.15 0
change - x + x + x
equilibrium (0.45 – x) (0.15 + x) x
Therefore, we have,
Ka = [CH3COO-][H3O+]/[CH3COOH] = (0.15 + x)(x)/(0.45 – x)
or, 1.8*10-5 = (0.15 + x)(x)/(0.45 – x)
or, 1.8*10-5.(0.45 – x) = (0.15 + x)(x)
or, 8.1*10-6 – 1.8*10-5x = 0.15x + x2
We have to assume that x is very small so that x2 can be neglected; therefor,
8.1*10-6 = 0.14998x
or, x = 5.40*10-5
Hence, [H3O+] = 5.40*10-5 M and pH = -log10[H3O+] = - log10(5.40*10-5) = 4.27
Fourth entry
We follow the same approach as above and obtain,
Ka = (0.45 + x)(x)/(0.15 – x)
or, 1.8*10-5.(0.15 – x) = 0.45x (small x approximation)
or, 2.7*10-6 – 1.8*10-5x = 0.45x
or, 2.7*10-6 = 0.44998 x
or, x = 6.00*10-6
and pH = -log10(6.00*10-6) = 5.22
Fifth entry
Follow same process as second and fourth entry and we get,
1.8*10-5.(0.30 – x) = 0.30x (small x approximation)
or, 5.4*10-6 = 0.29998x
or, x = 1.80*10-5
so that pH = -log(1.80*10-5) = 4.74
For the second entry in your table, we need to consider the reaction of acetate with water.
CH3COO- + H2O ------> CH3COOH + OH-
Since, OH- is generated, we must find out the pOH first. Also, Ka for acetic acid is give, so we can calculate Kb (dissociation constant for conjugate base) as
Ka.Kb = Kw where Kw is the ionic product for water = 10-14.
Therefore, Kb = 10-14/Ka = 10-14/(1.8*10-5) = 5.55*10-10
Kb = [CH3COOH][OH-]/[CH3COO-]
We use the approximation we used for the first entry, i.e, concentration of acetate is 0.6 M and the concentration of acetic acid and OH- are same (say c1).
Hence,
5.55*10-10 = (c1)(c1)/0.6
or, c12 = 5.55*10-10*0.6
or, c1 = (5.55*10-10)(0.6) = 1.825*10-5
The concentration of OH- IS 1.825*10-5 M and pOH = - log10[OH-] =
-log10(1.825*10-5) = 4.74
Therefore, pH = 14 – pOH (since pH + pOH = 14) = 14 – 4.74 = 9.26
Now, we can calculate the percentage error from the reported values as
% error = (obtained value – reported value)/(reported value) * 100
pH value reported
pH value obtained
% Error
2.25
2.48
10.22
3.86
4.27
10.36
8.83
9.26
4.87
5.40
5.22
(- 3.33) (negative value is because obtained value is less than reported value)
4.60
4.74
3.04
Volume of 0.6 M acetic acid added (mL)
Volume of 0.6 M acetate added (mL)
Total volume of solution (mL)
Concentration of acetic acid in solution (M)
Concentration of acetate in solution (M)
30
10
40
0.45
0.15
10
30
40
0.15
0.45
20
20
40
0.30
0.30
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