Question

Consider the titration of a 26.0?mL sample of 0.175M CH3NH2 with 0.155M HBr . Determine each of the following.

A. the initial pH

B. the volume of added acid required to reach the equivalence point

C.the pH at 5.0mL of added acid

D. the pH at 1/2 equivalence point

E. the pH at equivalence point

F.the pH after adding 4.0mL of acid beyond the equivalence point

Any help is appreciated!!! I'm just so lost with this one.

Answer 1

Q.Consider the titration of a 23.0 mL sample of 0.180 M CH3NH2 with 0.145 M HBr. Determine each of the following.
The initial pH is 11.95 and the volume of added acid required to reach the equivalence point is 28.6 mL.


1) The pH at 4.0 mL of added acid

2) the pH at one-half of the equivalence point

3) the pH at the equivalence point

4) the pH after adding 5.0 mL o acid beyond the equivalence point.

Answer:

CH3NH2 + H2O <=> CH3NH3+ + OH-
pOH = 14 - 11.95=2.05
[OH-]= [CH3NH3+]= 10^-2.05 =0.00891 M
[CH3NH2]= 0.180 - 0.00891=0.171 M
Kb = (0.00891)^2 / 0.171=4.64 x 10^-4
pKb = 3.33

1)
moles CH3NH2 = 0.0230 x 0.180=0.00414
moles HBr = 0.0040 L x 0.145 M=0.00058
CH3NH2 + H+ => CH3NH3+
moles CH3NH2 = 0.00414 - 0.00058=0.00356
moles CH3NH3+ = 0.00058
pOH = 3.33 + log 0.00058/ 0.00356=2.54
pH = 14 - 2.54=11.46

2)
moles HBr required to reach the half equivalence point = 0.00414/2 = 0.00207
moles CH3NH3+ formed = 0.00207
pOH = 3.33 + log 0.00207/0.00207= 3.33
pH = 10.67

3)
moles HBr required to reach the equivalence point = 0.00414
Volume HBr = 0.00414 / 0.145 M=0.0286 L
total volume = 0.0286 + 0.0230= 0.0516 L
moles CH3NH3+ = 0.00414
[CH3NH3+]= 0.00414 / 0.0516=0.0802 M
CH3NH3+ <=> CH3NH2 + H+
Ka = Kw/Kb = 1.0 x 10^-14 /4.64 x 10^-4 = 2.16 x 10^-11 = x^2 / 0.0802-x
x = [H+]= 1.32 x 10^-6 M
pH = 5.88

4)
volume HBr added = 0.0286 + 0.00500 L=0.0336 L
moles HBr = 0.0336 L x 0.145 M=0.00487
moles HBr in excess = 0.00487 - 0.00414 = 0.000730
[H+]= 0.000730/ 0.0336 =0.0217 M
pH = 1.66