Consider the titration of a 24.0 mL sample of 0.170 M CH3NH2 with 0.155 M HBr. D

ID: 965488 • Letter: C

Question

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

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