# Titration curves -- equivalence point vs. half equivalence point

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#### Revilla

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Am I right in thinking the 1/2 equivalence pt. is the region where the concentration of the two solutions are equal? If so, then what is the equivalence pt. because I always thought that was the region where the concentrations were equal.

#### RySerr21

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Am I right in thinking the 1/2 equivalence pt. is the region where the concentration of the two solutions are equal? If so, then what is the equivalence pt. because I always thought that was the region where the concentrations were equal.

say you are titrating a base with an acid...the equivalence point is when the amount of acid added is equal to the amount of base present.

edit: the 1/2 equivalence point is exactly what it sounds like. it is the point where the volume added is half of what it will be at the equivalence point. At this point, the pH = pKa. i think that point is important b/c its when the concentrations of base and acid are equimolar (i think..someonce can correct me)

#### Revilla

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So can you just divide the EP by 2 to get the 1/2 EP then? And on a titration curve, the 1/2 EP is in the middle of the first horizontal line, right?

#### bluemonkey

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The half-equivalence point is when half of your starting species has been either protonated (in the case of a base) or deprotonated (in the case of an acid). Say you have an acid to titrate, HA. At the half equivalence point, half of this acid has been deprotonated and half is still in its protonated form. This means that [HA]=[A-]. You can easily get the pH of the solution at this point via the HH equation, pH=pKa+log[A-]/[HA]. Since [A-]=[HA] at the half-eq point, the pH is equal to the pKa of your acid. If you look at a titration curve, your half-eq point is the flattest part of the cure while the equivalence point is at the steepest part.

I hope this helps!

#### RySerr21

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So can you just divide the EP by 2 to get the 1/2 EP then? And on a titration curve, the 1/2 EP is in the middle of the first horizontal line, right?

i dont see why dividing the EP by 2 wouldn't give you the 1/2 EP.

yes its in the middle of that first horizontal line. that is when the solution is acting like a buffer, which is why there isn't any change in pH

#### Revilla

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Thank you to both of you! That helps a lot!

#### lovebes

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[If this is by what you call thread hijacking, I apologize. I didn't mean to; thought it was convenient I post here rather than create a new one.]

Hey guys, a variation on the subject would be the distinction of the graphs when you have a strong on strong titration(eg strong acid via strong base titrant) versus a weak acid on strong base titrant. The graphs are different.

Reason is that there is a buffer point, which is defined as the "half-equivalence" point, where [HA] = [A-] (acid conc. = conj. base conc.). So, you'd see a plateau before graph intensely goes up to reach equivalence point.

See the weak acid/ strong titrant clearly shows this and it's noted in textbooks. What I'm wondering is ... wouldn't that point exist in the strong/strong titration as well? Wouldn't the graph have to go through a point where [HA] = [A-]?

I know that the fact that strong acids fully dissociate has something to do with this but can't be sure. My reason is that for weak acids, they only partially dissociate into A- and H+. Strong acids, it's already 100% H+ and A- initially, so [HA] is zero, thus can't have the point where [HA] = [A-].

Is my reasoning correct?

Seungjin

#### unique135

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[See the weak acid/ strong titrant clearly shows this and it's noted in textbooks. What I'm wondering is ... wouldn't that point exist in the strong/strong titration as well? Wouldn't the graph have to go through a point where [HA] = [A-]?

I know that the fact that strong acids fully dissociate has something to do with this but can't be sure. My reason is that for weak acids, they only partially dissociate into A- and H+. Strong acids, it's already 100% H+ and A- initially, so [HA] is zero, thus can't have the point where [HA] = [A-].
Seungjin

By definition, yeah. However, you won't see half-equivalent point on the graph (which plots pH over the function of strong base volume). If assuming that strong acids dissociate completely, strong acids must have passed through this point even before base is added. Here, the addition of base doesn't not drive strong acid dissociation.

#### Enjoy Life

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So if you are trying to solve for pH for a strong-weak titration you use the 1/2 equivalence pt... I guess my question is when do you use the HH equation for a strong-weak titration?

#### prometheusk

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The half-equivalence point is when half of your starting species has been either protonated (in the case of a base) or deprotonated (in the case of an acid). Say you have an acid to titrate, HA. At the half equivalence point, half of this acid has been deprotonated and half is still in its protonated form. This means that [HA]=[A-]. You can easily get the pH of the solution at this point via the HH equation, pH=pKa+log[A-]/[HA]. Since [A-]=[HA] at the half-eq point, the pH is equal to the pKa of your acid. If you look at a titration curve, your half-eq point is the flattest part of the cure while the equivalence point is at the steepest part.

I hope this helps!

Well put, I've always known that at the 1/2 equivilance point pH=PKa but never knew why. Why can't the books explain it like that, sweet and to the point.

#### dlouis

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Well put, I've always known that at the 1/2 equivilance point pH=PKa but never knew why. Why can't the books explain it like that, sweet and to the point.

But don't forget that for strong acid v strong base titration, the equivalence point is where HA = A-, not the half equivalence point and it always equals pH = pKa = 7

#### Rabolisk

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But don't forget that for strong acid v strong base titration, the equivalence point is where HA = A-, not the half equivalence point and it always equals pH = pKa = 7

This is not true. The equivalence point is where the formal concentration of HA (strong acid) equals the formal concentration of BOH- (strong base) added. They also react completely (neutralize) to produce a neutral solution. At the equivalence point, you basically have just H2O. pH = 7 is correct, but pKa = 7?? What species' pKa is 7 here? Certainly not water, the strong acid (pKa < -2), the protonated strong base (pKa > 16), or H3O+ (pKa &#8776; -2).

At the half equivalence point [HA] = [A-] is still true, although [HA] is really [H3O+], because a defining characteristic of a strong acid is that it is stronger than the hydronium ion, which is the strongest acid that can exist in an aqueous solution. Whether you believe it or not, there is still greatest buffering in this region, it's just that the pH is usually so low that it is of no practical use. You also can't use the HH equation.

#### dlouis

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This is not true. The equivalence point is where the formal concentration of HA (strong acid) equals the formal concentration of BOH- (strong base) added. They also react completely (neutralize) to produce a neutral solution. At the equivalence point, you basically have just H2O. pH = 7 is correct, but pKa = 7?? What species' pKa is 7 here? Certainly not water, the strong acid (pKa < -2), the protonated strong base (pKa > 16), or H3O+ (pKa &#8776; -2).

At the half equivalence point [HA] = [A-] is still true, although [HA] is really [H3O+], because a defining characteristic of a strong acid is that it is stronger than the hydronium ion, which is the strongest acid that can exist in an aqueous solution. Whether you believe it or not, there is still greatest buffering in this region, it's just that the pH is usually so low that it is of no practical use. You also can't use the HH equation.

whoops. Thank makes a lot of sense. Whenever I looked at a strong acid/base titration, the graph never showed the 1/2 equivalence point and it always showed H=A- at the equivalence point.

#### Joker88

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is this right?

the equivalence point for a weak base/acid with a strong base/acid just: Moles of titrant=moles of what you're titrating

1/2 eq point: 1/2 volume needed to neutrilize acid/base. where the pka of the acid is equal to the pH of the titrant. Concentration of HA= concentrate of A- (vice versa)

End point: above the equivalence point. Excess indicator used.

#### indianjatt

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is this right?

the equivalence point for a weak base/acid with a strong base/acid just: Moles of titrant=moles of what you're titrating

1/2 eq point: 1/2 volume needed to neutrilize acid/base. where the pka of the acid is equal to the pH of the titrant. Concentration of HA= concentrate of A- (vice versa)

End point: above the equivalence point. Excess indicator used.

You're absolutely correct about the equivalence point. Just keep in mind that the equivalence point for a weak acid/strong base will be >7 pH at 25 degrees C. Vice versa for a weak base/ strong acid.

I'm not sure what you mean in the bolded part, but here's an example. You have 50 mmol NH4+, so at the 1/2 equivalence point you will have 25 mmol OH-.

NH4+ + OH- <-----> NH3 + H2O
50 mmol 25 mmol
- 25 mmol - 25 mmol + 25 mmol

pH = pKa + log (25 mmol/25 mmol) = pKa. so pH = pKa for weak acid/base titrations.

#### Joker88

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if the 1/2 equivalence point represents the concentration of acid=concentration of conj base...what does the equivalence point for a weak acid/strong base titration represent? Moles of titrant=moles of titrated?

for equivalence point in a strong acid/strong base its where the acid is equal to the conjugative base right?

#### Eastmall

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high guys can anyone help me solve this problem? Thanks

Calculate the pH of a buffer made by combining 1.35 mol H3PO4 and 2.95 mol NaOH in 0.50L of aqueous solution. pKa1 = 2.12 pKa2 = 6.86 pKa3 = 12.4

#### tn4596

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is the ph 11.667...I am not totally sure but it goes like this:
adding 2.95 mol of NaOH in to 1.35 mol of H3PO4, after the first 2 dissociations, you are left with (2.95 - (1.35x2))= 0.25 mol NaOH and 1.35 mol HPO4.
We have for the 3rd dissociation: H + PO4 = HPO4
So, I guess let .25 mol NaOH = .25 mol H = .25 mol PO4. Then [A] = (.25mol/.5L) = .5 M
and [HA] = 2.7M
pH = 12.4 + log (.5M/2.7M) = 11.667

been a while since I've done this..might be totally wrong lol...