[webassignbuddy]

I figured out A.

K_b = y²/C_obase
4.96 x 10^-8 = y²/(0.1180)
7.65 x 10^-5 = y = [OH-]

pOH = -log [OH-]
pOH = 4.116

pH = 14 - pOH
pH = 14 - 4.116
pH = 9.88 (CORRECT)

BUT, how do I solve for b and c??

[Borek]

Where is the problem with b? It is a simple stoichiometry.

c - and the endpoint you have just a solution of the salt (which is a product of the neutralization). Proceed as if you were calculating pH of the salt, how you got there doesn't matter.

[webassignbuddy]

Where is the problem with b? It is a simple stoichiometry.

c - and the endpoint you have just a solution of the salt (which is a product of the neutralization). Proceed as if you were calculating pH of the salt, how you got there doesn't matter.

I understand part b, where you just do C1V1 = C2V2.

But part c still confuses me because i know

pH = pKa + log(nb/na)

At the equivalence point, nb = na and that quation above just becomes pH = pKa.

But pKa isn't given....

[sjb]

But pKa isn't given....

You have Kb?

[webassignbuddy]

But pKa isn't given....

You have Kb?

But I still got it wrong somehow. The answer is supposed to be 4.01 but I got 6.69.

Ka x Kb = Kw
Ka = Kw/Kb
Ka = (1 x 10^-14)/(4.96 x 10^-8
Ka = 2.02 x 10^-7

pKa = -log Ka
pKa = -log (2.02 x 10^-7)
pKa = 6.69

[Borek]

I understand part b, where you just do C1V1 = C2V2.

It works here, but it is not a general method.

At the equivalence point, nb = na

No. That's at midpoint.

I told you how to find the answer - and it has nothing to do with the Henderson-Hasslbalch equation.

[webassignbuddy]

It works here, but it is not a general method.

Then what is a general method to solve for part b? I didn't know what you meant by simple stoichiometry.

No. That's at midpoint.

I told you how to find the answer - and it has nothing to do with the Henderson-Hasslbalch equation.

My book and teacher said that at the equivalence point, nb = na ()
Edit: oh wait nevermind I misread that part in the book.

But aside from that...I'm confused by what you you said (i put it in bold) "c - and the endpoint you have just a solution of the salt (which is a product of the neutralization). Proceed as if you were calculating pH of the salt, how you got there doesn't matter."

I'm not given an endpoint, but I found the volume needed to reach the endpoint in part c, which is 39.74 mL.

The salt youre referring to...is it HB?? or HCl.

Because the equation I wrote is: B^- + H₃O⁺  HB + H₂O

[Borek]

It works here, but it is not a general method.

Then what is a general method to solve for part b? I didn't know what you meant by simple stoichiometry.

Just follow the reaction equation: http://www.titrations.info/titration-calculation

But aside from that...I'm confused by what you you said (i put it in bold) "c - and the endpoint you have just a solution of the salt (which is a product of the neutralization). Proceed as if you were calculating pH of the salt, how you got there doesn't matter."

I'm not given an endpoint, but I found the volume needed to reach the endpoint in part c, which is 39.74 mL.

You don't have to be "given" endpoint. By definition endpoint is when you added stoichiometric amount of titrant. Stoichiometric meaning "exactly the amount required to react with the substance present in the solution".

The salt youre referring to...is it HB?? or HCl.

HCl is not a salt.

This is not much different from titration of ammonia:

NH₃ + HCl NH₄Cl

Do you see the salt here? Your amine most likely behaves the same way.

[webassignbuddy]

It works here, but it is not a general method.

Then what is a general method to solve for part b? I didn't know what you meant by simple stoichiometry.

Just follow the reaction equation: http://www.titrations.info/titration-calculation

But aside from that...I'm confused by what you you said (i put it in bold) "c - and the endpoint you have just a solution of the salt (which is a product of the neutralization). Proceed as if you were calculating pH of the salt, how you got there doesn't matter."

I'm not given an endpoint, but I found the volume needed to reach the endpoint in part c, which is 39.74 mL.

You don't have to be "given" endpoint. By definition endpoint is when you added stoichiometric amount of titrant. Stoichiometric meaning "exactly the amount required to react with the substance present in the solution".

The salt youre referring to...is it HB?? or HCl.

HCl is not a salt.

This is not much different from titration of ammonia:

NH₃ + HCl NH₄Cl

Do you see the salt here? Your amine most likely behaves the same way.

That's the thing though.
I have no idea where to begin to approach part c because this seems vague to me...

[Borek]

I have no idea where to begin to approach part c because this seems vague to me...

What seems vague to you?

Do you know how to calculate pH of NH₄Cl solution? Hint: it contains a weak acid.

[webassignbuddy]

I have no idea where to begin to approach part c because this seems vague to me...

What seems vague to you?

Do you know how to calculate pH of NH₄Cl solution? Hint: it contains a weak acid.

I figured it out... Maybe the whole salt concept (NH4Cl) was throwing me off but this is what I did.
I was more wondering about how to construct the table.

H₂O + HB  H₃O⁺ + B^-
------(~0)------(3.06)--(3.06)
------(+3.06)--(-3.06)--(-3.06)
------(3.06)----(~0)-----(~0)

C_o (HB) = mmol HB/total mL
C_o (HB) = 3.06/(26 mL + 39.74 mL of titrant added)
C_o = 3.06/65.7
C_o = 0.04657

Ka = x₂/C_o
(Ka x C_o)^(1/2) = x = [H₃O⁺]
((2.02 x 10^-7)(0.04657))^(1/2)  = x = [H₃O⁺]
9.6997 x 10^-5 = x = [H₃O⁺]

pH = -log [H₃O⁺]
pH = -log(9.6997 x 10^-5)
pH = 4.01

[webassignbuddy]

How do I solve for part f though?
The answer is supposed to be 1.11

I know I should construct a different table to visualize this better but I'm not sure what values should go in it.

[Borek]

H₂O + HB  H₃O⁺ + B^-
------(~0)------(3.06)--(3.06)
------(+3.06)--(-3.06)--(-3.06)
------(3.06)----(~0)-----(~0)

No idea what these numbers are, nor how the reaction is related to the problem. (That is - I think I know what you mean, but i you want use to help, don't make us guess, explain what you are doing, We are not mind readers.)

[Borek]

How do I solve for part f though?
The answer is supposed to be 1.11

When you add a huge excess of the titrant you end basically with just a titrant solution.

Try to calculate concentration of the acid if you add 26 mL of 0.1180 M base solution to 1 cubic meter of 0.0772 M HCl. Assume a base strong to make calculations easier.

[webassignbuddy]

How do I solve for part f though?
The answer is supposed to be 1.11

When you add a huge excess of the titrant you end basically with just a titrant solution.

Try to calculate concentration of the acid if you add 26 mL of 0.1180 M base solution to 1 cubic meter of 0.0772 M HCl. Assume a base strong to make calculations easier.

What is 1 cubic meter of 0.0772 M HCl? Is that 1,000,000 mL of 0.0772 M HCl?

Oh ok!
I got it right!
Thanks!!