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fodog414 fodog414
wrote...
Posts: 279
9 years ago Edited: 9 years ago, fodog414
Consider the titration of 100.0 mL of 0.500 M CH3NH2 by 0.250 M HCl.
(Kb for CH3NH2 = 4.4×10-4)

Part 1
Calculate the pH after 0.0 mL of HCl added.

pH =

Part 2
Calculate the pH after 25.0 mL of HCl added.

pH =

Part 3
Calculate the pH after 80.0 mL of HCl added.

pH =

Part 4
Calculate the pH at the equivalence point.

pH =

Part 5
Calculate the pH after 300.0 mL of HCl added.

pH =

Part 6
At what volume of HCl added, does the pH = 10.64?

We never went over this, but if someone could explain a few or if they would like all of them it should be straight forward with a thorough explanation. Its due tomorrow  :-\so any help is truly appreciated.

Thank you!
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wrote...
9 years ago
Consider the titration of 100.0 mL of 0.500 M CH3NH2 by 0.250 M HCl.

(Kb for CH3NH2 = 4.4 x 10^-4)
At what volume of HCl Added does the pH=10.64

[OH-] = sqrt(Kb*0.5) = 0.01483 M

initial milli-moles of OH- = (0.01483*100) = 1.483

when pH = 10.64,
pOH = 14-10.64 = 3.36
final [OH-] = 4.365x10^-4 M

let V ml of 0.25 M HCl be added

total volume of solution = (100+V) ml

final milli-moles of
OH- = (100+V)*(4.365x10^-4)

change in milli-moles of OH- = 1.483-[(100+V)*(4.365x10^-4)] = [1.44-(4.365x10^-4)*V]

this must be equal to milli-moles of H+ added which is equal to 0.25V

hence,

0.25V = [1.44-(4.365x10^-4)*V]

solving, V = 5.75 ml
so, pH becomes 10.64 when 5.75 ml acid is added

Consider the titration of 100.0mL of 0.500 M NH₃ (Kb = 1.8 x 10^-5) with a 0.500 M HCl. After 50.0 mL HCl have been added, the [H₃0] of the solution is:

Original moles of NH3 = 100/1000 x 0.500 = 0.05 mol

Moles of HCl added = 50/1000 x 0.500 = 0.025 mol

HCl + NH3 => NH4Cl

After addition of HCl,

Moles of NH3 left = 0.05 - 0.025 = 0.025 mol

Moles of NH4Cl formed = 0.025 mol

Kb(NH3) = 1.8 x 10-5 => Ka(NH4Cl) = Kw/Kb(NH3)

= 10-14/1.8 x 10-5 = 5.556 x 10-10

pH = pKa + log ([NH3]/[NH4Cl])

= -log (5.556 x 10-10) + log [(0.025/0.150)/(0.025/0.150)]

= 9.255

pH = -log [H3O+] = 9.255

[H3O+] = 10-9.255 = 5.556 x 10-10 M
fodog414 Author
wrote...
9 years ago
Thank you so much for the quick reply!

So following your format

for part 2 i got 10.49

part 3 10.46

but for part 5 i hit a problem i got -.025 finding what is left .05-.075? Should this be reversed?

and you don't have to say but is part 1-4 correct?

wrote...
9 years ago
Tbh, I didn't answer these myself, I was hoping you'd know what to do with them. As far as I know, they might even be wrong Frowning Face
fodog414 Author
wrote...
9 years ago
Yeah they were all incorrect Frowning Face
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