TBR titrations

[californiamed]

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Titrations is making me feel dumb.

On the second page of the TBR titrations, it says: "in the first system, Weak acid is in extreme excess, relative to its conj base":

Initially [A-]/[HA]=1/199 ----> after addition of OH-: [A-]/[HA] = 2/998.

Why does [A-] increase? It says the addition of a strong base or acid shifts the ratio of weak acid to its conjugate base. WHY????? and why by 2? I'm confused 🙁

 

[LoLCareerGoals]

Titrations is making me feel dumb.

On the second page of the TBR titrations, it says: "in the first system, Weak acid is in extreme excess, relative to its conj base":

Initially [A-]/[HA]=1/199 ----> after addition of OH-: [A-]/[HA] = 2/998.

Why does [A-] increase? It says the addition of a strong base or acid??? shifts the ratio of weak acid to its conjugate base. WHY????? and why by 2? I'm confused 🙁

Last time I checked 1/199 is bigger than 2/998, so I am not sure how you are getting that [A-] increases (unless [HA] increases even more) from those fractions. Did you mean 1/999 -> 2/998? I don't have this book.

Now as for the concept:
[HA] <-> [H+] + [A-] where [A-]/[HA] = 1/999. Solution is slightly acidic due to few [H+] floating around. If we add a strong base i.e. OH-s, then it will react with H+s lowering their concetration. If [H+] concentration is lowered the balance in the reaction will shift right! Shifting right it produces more [A-] by dissociating [HA] to make them. Hence [A-] goes up and [HA] goes down. As for amount by which it goes up/down depends on amound of OH- added.
As for "or acid" above, that makes no sense to me. Strong acid will make more [H+] and shift the balance to the left -> dropping [A-] and making more [HA].
Edit: Nm, it just says shifts, so statement is fine. Base shifts it one way, acid the other, disregard my "???".

 

[californiamed]

Last time I checked 1/199 is bigger than 2/998, so I am not sure how you are getting that [A-] increases (unless [HA] increases even more) from those fractions. Did you mean 1/999 -> 2/998? I don't have this book.

Now as for the concept:
[HA] <-> [H+] + [A-] where [A-]/[HA] = 1/999. Solution is slightly acidic due to few [H+] floating around. If we add a strong base i.e. OH-s, then it will react with H+s lowering their concetration. If [H+] concentration is lowered the balance in the reaction will shift right! Shifting right it produces more [A-] by dissociating [HA] to make them. Hence [A-] goes up and [HA] goes down. As for amount by which it goes up/down depends on amound of OH- added.
As for "or acid" above, that makes no sense to me. Strong acid will make more [H+] and shift the balance to the left -> dropping [A-] and making more [HA].
Edit: Nm, it just says shifts, so statement is fine. Base shifts it one way, acid the other, disregard my "???".

Thanks for the explanation, I get it now. I'm now confused on another example about equivalences:

Which of the following solutions results in a buffer with a pH of 5, given that HA has a pka of 4.7?
A) HA w/ one-half equivalent A-
B) A0 w/ one equivalent of HA
C) HA w/ one-third equivalent of OH-
D) A- w/ one-third equiv of H3O+

answer - D
-I know equivalence pnt = when moles of acid = moles of titrant (base). But the answer also mentions something about how in choice D, for ex, "one third of an equiv of HA forms from the reaction, and 2/3 of equi of A- is left over". Why is HA formed and A- left over? What does that mean? Thank you 🙂

 

[LoLCareerGoals]

Remember that A- is a base (conj of a weak acid). Give a base some H30+ and it will make water and conj acid. So If [H3O+]/[A-] = 1/3, then [H3O+] + [A-] -> H2O + HA with 2/3 left over since there isn't enough H3O+ to react with all of the A- (H3O+ is a limiting reactant).
Edit: Also I know you wanted a quick answer, but if it is not too much trouble could you put full questions, so people can practice and make their own mistakes?

 

[californiamed]

Thanks, i edited the question... i'm still confused at what you mean by " 2/3 left over since there isn't enough H3O+ to react with all of the A- (H3O+ is a limiting reactant)." What does one-third equiv of H3O mean? Less H30 to start w/? And book answer says "one third of an equiv of HA forms from the reaction, and 2/3 of equivalent of A- is left over".. So 1/3 H30 equivalent means one third of HA forms? So why does this affect A-?

 

[LoLCareerGoals]

Say [A-] = x. What they are saying is that [H3O+] = x/3 = [H+].
Let's look at this equation:

HA + H20 <-> [H3O+] + [A-] or simply: HA <-> [H+] + [A-] (remember H+ and H3O+ are used interchangably).

"So why does this affect A-? "
It takes [A-] to make [HA].

What it comes down to I think is the equation for PH of a buffer: pKa + log10(base/acid) = 4.7 + log10(3/1) = 5.17.

 

[californiamed]

thanks.

also, what does it mean when it say "adding half of an equivalent of NaOH" to a NaHCO3 converts half of the HCO3- to its conjugate base CO3-... What's half of an equivalent?

Another example (non passage) - adding an equivalent of H2Co3 to asodium bicarbonate solution results in equal proportions of HCO3- and its conjugate acid H2CO3.

 

[sc4s2cg]

That just means if you have 50mL of 5Molarity X, than add 25mL of 5Molarity Y.

Half of an equivalent means the same concentration, but half the volume.

 

[californiamed]

Thanks everyone.

And lastly, for the henderson-hasselbach equation, for the [A-]/[HA] part, is that CONCENTRATION of each? In TBR, they showed a table (table 5.3, page 306) where the [A-]/[HA] part was in mL -- i.e. mlOH-/(50 - mLOH-). You can't put liters in the equation, it has to be concentration -- i.e. - molarity (something divided by volume), right?

 

[sc4s2cg]

Yes, it is concentrations. The [] signifies concentration.

 

[californiamed]

weird, anyone knows why tbr decided to have the concnetration in mL instead of molarity?