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Half equivalence point

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Hemichordate

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When it says that the PH = pKa at the half equivalence point, is it saying that the acid species (ie HCl) has that pKa or that the base species (Cl-) has that pKa as well?
 

Salient

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Edit: Totally misunderstood your question because it was so short and ranted on about stuff I'm sure you already know.
 
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ThirdEye

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When it says that the PH = pKa at the half equivalence point, is it saying that the acid species (ie HCl) has that pKa or that the base species (Cl-) has that pKa as well?

If you're titrating with a base, then it's the acid.

If you're titrating with an acid, then it's the base.
 

Salient

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So if you're titrating a base with an acid, would you still say "pKa of base" or would you need to determine the pKb?

Can you give an example problem? I'm going to bed but I'll check it first thing in the morning. pKa and pKb are interchangeable, but you shouldn't ever need to convert a pKa into a pKb to solve a problem.
 

MCAT guy

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So if you're titrating a base with an acid, would you still say "pKa of base" or would you need to determine the pKb?

You would say pKb... the whole point of the Ka (acid dissociation constant) is to tell us how much acid dissociates in water. pKa is just an easier way to write it. To say the pKa of the base wouldn't make much sense BUT pKa + pKb = 14, so if you have one you have the other.
 

MCAT guy

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It's still pKa for titrating a base when we're dealing with pH. Only if they specify pOH, will it be pKb.

right. you use pKa when you want to find the pH and pKb when you want to find pOH.

Remember no matter what they give you, you can find the pKb or pKa because pKb + pKa = 14
 

Isoprop

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it's the pKa of the sample, which is a weak acid. your question doesn't make sense because there is no such thing as a pKa of a base. You can find the pKa of the conjugate acid of a base, but not the base itself.

Also, nobody talks about the pKa of HCl because HCl is a strong acid. It dissociates 100% for the MCAT unless told otherwise.

Example: titrating HF with NaOH.

At the half equivalence point, the pH = pKa of the HF.
 

Hemichordate

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I guess since I asked a confusing question I'll give an example:

So if I have a weak acid (i.e. acetic acid) and I titrate it with a base, when I add enough base, it'll reach the half equivalence point, at which point the concentration of the weak acid and its conjugate base will be equal. The pH would also be equal to the pKa at this point.

At this point, could I say that this pKa value is the pKa of the original weak acid? Could I also say that this same pKa is also the pKa of the conjugate base?

I guess when people say sample, I'm just trying to figure out which one it is (acid or conjugate base), or if it's both.
 

Fletcher Moore

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pKa is a constant that is associated with any given acid. It does not change. It is actually associated with an equilibrium equation HA -> H+ + A-.

Eg. acetic acid is 4.76. Always.

That means at pH 4.76, the acid will be half exploded, as I like to consider it.

pH is associated with an entire solution and pH changes because it is a measure of how many free H+ ions are in solution.

At a higher pH, there are less free H+, so the acid has more room to stretch and ejects some more H+. At a lower pH there are more free H+ in the solution so acid is forced back together.
 

Salient

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Right. So when you're titrating acetic acid with a strong base, you are doing something like:

CH3COOH + NaOH ---> CH3COO- + H2O

At the half equivalence point, you've used up half of the original acetic acid, and the concentrations of the acetic acid and its conjugate base are equal.

You now have an equilibrium in the solution where Acetic acid and acetate are both reacting with water to create each other.

CH3COOH + H2O <----> CH3COO- + H3O+

The Ka for this equation is [CH3COO-][H3O+]/[CH3COOH].

At the half-equivalance point, [CH3COO-] is equal to [CH3COOH], so they cancel out of the equation and Ka = [H3O+]

take the negative log of both sides and you have pKa = pH.

You could do the exact same problem with pKb, if you REALLY wanted:

If we write our equilibrium as:
CH3COO- + H2O <----> CH3COOH + OH-

then Kb = [CH3COOH][OH-]/[CH3COO-]

again at the half-equivalence point, the concentrations are the same and pKb = pOH.

Understand that the pKa and pKb terms refer to the ratio of both conjugates as well as the H3O+ or OH- concentration. So the pKa isn't a property of ONLY the conjugate acid or base, it's a property shared by both.

Kb just refers to the ratio of products you would find in a solution if you dumped a bunch of conjugate base into water (e.g. NaCH3COO- ) while Ka refers to the ratio of products you would find in a solution if you dumped a bunch of conjugate acid into water (e.g. CH3COOH). They're two sides of the same coin.


NOW:

If you're titrating a weak base with a strong acid, you might have an equation like:

NH3 + HCl ---> NH4+ + Cl-

So before completion, you have an equilibrium that looks like:

NH3 + H2O <----> NH4+ + OH-

and you can solve this using the pKb....

but you can re-write the equilibrium as:

NH4+ + H2O <----> NH3 + H3O+

and solve it using the pKa. Once again, the pKa and pKb BOTH refer to BOTH the concentrations of NH4+ and NH3.

All that said, if this comes up on the MCAT it will probably be a conceptual question like:

Aqueous ammonia (pKb 4.75) is titrated with HCl. At the half-equivalence point:

A. pH is < 4
B. pH is > 4 and < 7
C. pH is = 7
D. pH is > 7

when the two species are equal in concentration, pH = pKa and pOH = pKb. the answer is D.
 

narnia

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Right. So when you're titrating acetic acid with a strong base, you are doing something like:

CH3COOH + NaOH ---> CH3COO- + H2O

At the half equivalence point, you've used up half of the original acetic acid, and the concentrations of the acetic acid and its conjugate base are equal.

You now have an equilibrium in the solution where Acetic acid and acetate are both reacting with water to create each other.

CH3COOH + H2O <----> CH3COO- + H3O+

The Ka for this equation is [CH3COO-][H3O+]/[CH3COOH].

At the half-equivalance point, [CH3COO-] is equal to [CH3COOH], so they cancel out of the equation and Ka = [H3O+]

take the negative log of both sides and you have pKa = pH.

You could do the exact same problem with pKb, if you REALLY wanted:

If we write our equilibrium as:
CH3COO- + H2O <----> CH3COOH + OH-

then Kb = [CH3COOH][OH-]/[CH3COO-]

again at the half-equivalence point, the concentrations are the same and pKb = pOH.

Understand that the pKa and pKb terms refer to the ratio of both conjugates as well as the H3O+ or OH- concentration. So the pKa isn't a property of ONLY the conjugate acid or base, it's a property shared by both.

Kb just refers to the ratio of products you would find in a solution if you dumped a bunch of conjugate base into water (e.g. NaCH3COO- ) while Ka refers to the ratio of products you would find in a solution if you dumped a bunch of conjugate acid into water (e.g. CH3COOH). They're two sides of the same coin.


NOW:

If you're titrating a weak base with a strong acid, you might have an equation like:

NH3 + HCl ---> NH4+ + Cl-

So before completion, you have an equilibrium that looks like:

NH3 + H2O <----> NH4+ + OH-

and you can solve this using the pKb....

but you can re-write the equilibrium as:

NH4+ + H2O <----> NH3 + H3O+

and solve it using the pKa. Once again, the pKa and pKb BOTH refer to BOTH the concentrations of NH4+ and NH3.

All that said, if this comes up on the MCAT it will probably be a conceptual question like:

Aqueous ammonia (pKb 4.75) is titrated with HCl. At the half-equivalence point:

A. pH is < 4
B. pH is > 4 and < 7
C. pH is = 7
D. pH is > 7

when the two species are equal in concentration, pH = pKa and pOH = pKb. the answer is D.


Argh I'm so close to understanding this, but online there are so many conflicting or confusing descriptions!

So in your second example, adding HCl to weak base NH3, would it be at the half-equivalence point or at the equivalence point that [B-] = [BH] ? And then what would be the relationship between the pH, pKa, and pKb at that point?
 
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