In: Chemistry
Consider the titration of a 24.0 −mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.
Part A the initial pH Express your answer using two decimal places. pH =
Part B the volume of added acid required to reach the equivalence point V = mL
Part C the pH at 6.0 mL of added acid Express your answer using two decimal places. pH =
Part D the pH at one-half of the equivalence point Express your answer using two decimal places. pH =
Part E the pH at the equivalence point Express your answer using two decimal places. pH =
Part F the pH after adding 6.0 mL of acid beyond the equivalence point
A )
Kb = 5.0 x 10^-4
CH3NH2 + H2O --------------------CH3NH3+ + OH-
0.170 -x x x
Kb = x^2 / 0.170-x
5 x 10^-4 = x^2 / 0.170-x
x = 8.97 x 10^-3
[OH-] = 8.97 x 10^-3 M
pOH = -log [OH-] = 2.05
pH + pOH = 14
pH = 11.95
part B)
volume at equivalence point = 24 x 0.170 / 0.155 = 26.3 mL
part C)
millimoles of base = 24 x 0.170 = 4.08
millimoles of acid = 0.155 x 6 =0.93
CH3NH2 + HBr-----------------> CH3NH3+Br
4.08 0.93 0
3.15 0 0.93
pOH = pKb + log(salt /base)
pOH = 3.30 + log (0.93 / 3.15)
pOH = 2.77
pH = 11.23
par D ) at half equivalenc epoint pOH = pKb
pOH = 3.30
pH + POH = 14
pH = 10.70