Calculate the pH at the equivalence point for the titration of 0.20 M aniline (C6H5NH₂, Kb = 3.8 x 10-10) with 0.20 M HCI. [Hint: remember the dilution factor caused by addition of titrant to the aniline solution.] 2.79 C₂H5NH₂ + HCl -> C₁ H 5 NH ₂ + 1² H₂6 + RNH₂ →RNH₂ + H₂0² 0.10M A) B) C) D) 5.58 5.21 8.79 I C E 0.10.x K₁ = 1₁0x10-14 3.8/0.10 E)None of the above. = 2.6x10'5 pH = -lay (1-²x10²4)= 2.79 TX 0.10.x x²= 2.1x10²

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter15: Additional Aqueous Equilibria
Section: Chapter Questions
Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...
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3
Calculate the pH at the equivalence point for the titration of 0.20 M aniline
(C6H5NH2, Kb = 3.8 x 10-10) with 0.20 M HCI. [Hint: remember the dilution factor
caused by addition of titrant to the aniline solution.]
2.79
CHSNH, HOC NH
H₂6 + RAH₂ RNH₂ + H₂0+
->
3
0.10M
XA)
B)
с
5.58
5.21
8.79
I
C
E
0.10.x
K₁ = 1₁0×10-14
3.8x/0.10
E)None of the above.
41°
= 2.6x10's-
X
x²= 2.1x10²
pH = -107 (1²x10²8) = 2.79
0.10.x
-4
Transcribed Image Text:个 3 Calculate the pH at the equivalence point for the titration of 0.20 M aniline (C6H5NH2, Kb = 3.8 x 10-10) with 0.20 M HCI. [Hint: remember the dilution factor caused by addition of titrant to the aniline solution.] 2.79 CHSNH, HOC NH H₂6 + RAH₂ RNH₂ + H₂0+ -> 3 0.10M XA) B) с 5.58 5.21 8.79 I C E 0.10.x K₁ = 1₁0×10-14 3.8x/0.10 E)None of the above. 41° = 2.6x10's- X x²= 2.1x10² pH = -107 (1²x10²8) = 2.79 0.10.x -4
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