a. titration of 27.9 mL of a solution of the weak base aniline, C 6 H 5 NH 2 , r

ID: 504481 • Letter: A

Question

a. titration of 27.9 mL of a solution of the weak base aniline, C6H5NH2, requires 28.64 mL of 0.160 M HCl to reach the equivalence point.

C6H5NH2(aq) + H3O+(aq) C6H5NH3+(aq) + H2O()

What was the concentration of aniline in the original solution?

[C6H5NH2] ------------------------- M

b. You titrate 25.0 mL of 0.60 M NH3 with 0.60 M HCl.

What is the pH of the NH3 solution before the titration begins?

Kb = 1.8×10-5

pH ------------------------------

a. titration of 27.9 mL of a solution of the weak base aniline, C6H5NH2, requires 28.64 mL of 0.160 M HCl to reach the equivalence point.

C6H5NH2(aq) + H3O+(aq) C6H5NH3+(aq) + H2O()

What was the concentration of aniline in the original solution?

[C6H5NH2] ------------------------- M

a) millimoles of base = millimoles of acid

27.9 x C = 28.64 x 0.160

C = 0.164 M

[C6H5NH2] = 0.164 M

pOH = 1/2 [pKb - log C]

pOH = 1/2 [4.74 -log 0.60]

pOH = 2.48

pH + pOH = 14

pH = 11.52

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