2. Assume that you have 20.00 mL of 0.1000 M dimethylamine (pKb = 3.230) in an E
ID: 511701 • Letter: 2
Question
2. Assume that you have 20.00 mL of 0.1000 M dimethylamine (pKb = 3.230) in an Erlenmeyer flask and you titrated with 0.15000 M strong acid or base. Answer the following questions.
a. Would you add a strong acid or a strong base to the buret? Explain why.
b. What is the initial pH prior to adding any strong acid or base? Perform calculation.
c. What is the pH at the half-equivalence point?
d. What is the pH at the equivalence point?
e. Draw the titration curve with pH on the y-axis and fraction of dimethylamine neutralized on the x-axis. Above the titration curve draw the fractional distribution plot ( on the y-axis and fraction of dimethylamine neutralized on the x-axis).
Explanation / Answer
(a)
We will be adding a strong acid because dimethylamine is a base and its neutralization can be done using an acid only.
(b)
Let BOH represent the base.
Using the following equation:
Kb = ([B+][OH-])/[BOH] = (x2)/(0.1-x) = 0.0005888
Solving we get:
[OH-] = x = 0.00738
So,
pOH = 2.13
So,
pH = 11.87
(c)
pOH at half equivalence point for a weak base is equal to its pKb value. This is because the conc. of base and its salt are the same at this point, so the logarithm term in Henderson Hasselbach equation evaluates to 0.
So, at half equivalence point, pOH = 3.23
So, pH = 14-pOH = 14-3.23 = 10.77
(d)
First we calculate the volume of acid required using:
M1V1 = M2V2
0.1*20 = 0.15*V2, so
V2 = 13.33 mL
So, conc of salt at equivalence point = moles of salt/total volume = (0.02*0.1)/(0.02+0.0133) = 0.06 M
Let AB represent the salt formed
At equivalence point, the following reaction takes place:
A- + B+ + H2O ----> BOH + A- + H+
Ka = ([BOH][H+])/[B+] = (x2)/(0.06-x)
So, [H+] = x
Also, Ka*Kb = 10-14
So, Ka = 1.69*10-11
Solving we get:
x = 10-6
So, pH = 6
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