Consider the titration of a 27.0 −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.
SubmitMy AnswersGive Up Part B the volume of added acid required to reach the equivalence point
SubmitMy AnswersGive Up Part C the pH at 6.0 mL of added acid Express your answer using two decimal places.
SubmitMy AnswersGive Up Part D the pH at one-half of the equivalence point Express your answer using two decimal places.
SubmitMy AnswersGive Up Part E the pH at the equivalence point Express your answer using two decimal places.
SubmitMy AnswersGive Up Part F the pH after adding 4.0 mL of acid beyond the equivalence point Express your answer using two decimal places.
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suppose base CH3NH2 = B
Kb of Base = 4.2 x 10^-4
pKb = -log Kb = -log (4.2 x 10^-4 ) = 3.38
(a) before addition of any HBr
B + H2O -------------------> BH+ + OH-
0.170 0 0 ---------------> initial
0.170-x x x --------------> equilibrium
Kb = [BH+][OH-]/[B]
4.2 x 10^-4 = x^2 / 0.170-x
x^2 + 4.2 x 10^-4 x - 7.14 x 10^-5 = 0
x = 8.24 x 10^-3
x= [OH-] = 8.24 x 10^-3M
pOH = -log[OH-]
pOH = -log (8.24 x 10^-3)
pOH = 2.08
pH + pOH = 14
pH = 11.92
(b) volume added equivalence point
millimoles of B = 27 x 0.170 = 4.59
millimoles of acid = V x 0.155 = 0.155 V
at equivaelce point
0.155 V = 4.59
V = 29.6 mL
volume = 29.6
(c) addition 6 ml acid
millimoles of acid = 6 x 0.155 = 0.93
B + H+ -------------------> BH+
4.59 0.93 0
3.66 0 0.93
pOH = pKb + log [0.93 / 3.66]
pOH = 3.38 +log [0.93 / 3.66]
pOH = 2.785
pH = 11.22
part D )
half equivalence point
pKb = pOH
pOH = 3.38
pH = 10.62
Consider the titration of a 27.0 −mL sample of 0.170 M CH3NH2 with 0.155 M HBr....
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