I really need these questions answered...

[BoneMental]

The only parts about general chemistry that really get me are super important.

I really would appreciate help answering these two questions.

1) Can you use the Henderson Hasselbalch equation for both titrations and buffer systems?

2) I don't fully understand, in a titration, when ph = pKA and when the equivalence point is reached. Although I know that the midpoint of the horizontalish line is where ph=pKA and that the midpoint of the vertical line is the equivalence point, I do not understand this in practicality. Can someone explain this to me in an example (ie, HCl and NaOH)?

Once again, I really appreciate the help. Ahhhhhh!🙁

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[ilovemcat]

The only parts about general chemistry that really get me are super important.

I really would appreciate help answering these two questions.

1) Can you use the Henderson Hasselbalch equation for both titrations and buffer systems?

2) I don't fully understand, in a titration, when ph = pKA and when the equivalence point is reached. Although I know that the midpoint of the horizontalish line is where ph=pKA and that the midpoint of the vertical line is the equivalence point, I do not understand this in practicality. Can someone explain this to me in an example (ie, HCl and NaOH)?

Once again, I really appreciate the help. Ahhhhhh!🙁

1.) HH equation is specifically used to describe a buffer system. An ideal buffer system has equal amounts of Weak Acid and CB, producing a 1:1 ratio.

3 Ways to make an ideal buffer system (example):

Weak Acid and Salt of Conjugate Base
5M HF and 5M NaF ---> This produces a 1:1 ratio (of HF and F-)

Weak Acid and 1/2 as much Strong Base
10M HF and 5M NaOH (or any strong base) --> This producea a 1:1 ratio (of HF and F-)

Salt of Conjugate Base and 1/2 as much Strong Acid
10M NaF and 5M HCl (or any strong acid) --> This produces a 1:1 ratio (of HF and F-)

2.) When pH = pKa you have an ideal buffer system. That's basically what the HH is useful for. If you have an equal concentration of weak acid and conjugate base, then your solution behaves as a buffer. Now if you keep adding Strong Acid or Strong Base, this 1:1 ratio will be disturbed - either producing more conjugate base (until all the HF has reacted in the presence of more NaOH) or producing more conjugate acid (until all F- has reacted in the presence of more HCl).


Just a note: A buffer doesn't exist for a strong acid or strong base because they react completely in the forward direction. For instance, HCl acid dissociates to H+ and Cl- but the reverse reaction doesn't happen. You don't see Cl- ions combining with H+ ions to produce HCl. This is why a buffer system is specifically a mixture of weak acid and conjugate base.

A second note is that in my example above, F- ions cannot just exist in a solution. You can't just dump an equimolar concentration F- ions into a mixture of HF to produce a buffer. The way F- ions are added to a solution is via a salt. Not just any salt though - you need to ensure that the F- ion is combined with a neutral cation / metal (one that isn't acidic). Examples of these are all the Conjugate Acids of Strong Bases like (Li+, Na+, K+, etc.). This is true for any situation in which you're trying to add the conjugate of a weak acid into a solution.

By the way, when pH = pKa, it's NOT the equivalence point that's reached. It's the HALF-equivalence point. They're two entirely different things. Half equivalence point is 1/2 the volume needed to neutralize the acid/base you are titrating. This is where we have a buffer region. Equivalence point is when moles of acid = moles of base.

 

[Rabolisk]

An excellent post against by ilovemcat, who is on fast track to a 40+. I would add that there is a fourth way of making a buffer, by combining a weak base with a salt of its conjugate acid. An example is NH3 and NH4Cl. HH equation can be used any time you are confident that you know [HA] and [A-], concentrations of an acid and its conjugate base, at equilibrium. Note that [HA] and [A-] at equilibrium is different from the formal concentrations. To clear up exactly what is meant by equivalence point, it is the point at which the moles of titrant added (strong acid or base) is equal to the initial moles of the analyte present (starting acid or base), resulting in HA being converted fully to A-. With polyprotic analytes, you have multiple equivalence points, with the first corresponding with H3A being converted fully to H2A-, for example.

 

[flodhi1]

1.) HH equation is specifically used to describe a buffer system. An ideal buffer system has equal amounts of Weak Acid and CB, producing a 1:1 ratio.

3 Ways to make an ideal buffer system (example):

Weak Acid and Salt of Conjugate Base
5M HF and 5M NaF ---> This produces a 1:1 ratio (of HF and F-)

Weak Acid and 1/2 as much Strong Base
10M HF and 5M NaOH (or any strong base) --> This producea a 1:1 ratio (of HF and F-)

Salt of Conjugate Base and 1/2 as much Strong Acid
10M NaF and 5M HCl (or any strong acid) --> This produces a 1:1 ratio (of HF and F-)

2.) When pH = pKa you have an ideal buffer system. That's basically what the HH is useful for. If you have an equal concentration of weak acid and conjugate base, then your solution behaves as a buffer. Now if you keep adding Strong Acid or Strong Base, this 1:1 ratio will be disturbed - either producing more conjugate base (until all the HF has reacted in the presence of more NaOH) or producing more conjugate acid (until all F- has reacted in the presence of more HCl).


Just a note: A buffer doesn't exist for a strong acid or strong base because they react completely in the forward direction. For instance, HCl acid dissociates to H+ and Cl- but the reverse reaction doesn't happen. You don't see Cl- ions combining with H+ ions to produce HCl. This is why a buffer system is specifically a mixture of weak acid and conjugate base.

A second note is that in my example above, F- ions cannot just exist in a solution. You can't just dump an equimolar concentration F- ions into a mixture of HF to produce a buffer. The way F- ions are added to a solution is via a salt. Not just any salt though - you need to ensure that the F- ion is combined with a neutral cation / metal (one that isn't acidic). Examples of these are all the Conjugate Acids of Strong Bases like (Li+, Na+, K+, etc.). This is true for any situation in which you're trying to add the conjugate of a weak acid into a solution.

By the way, when pH = pKa, it's NOT the equivalence point that's reached. It's the HALF-equivalence point. They're two entirely different things. Half equivalence point is 1/2 the volume needed to neutralize the acid/base you are titrating. This is where we have a buffer region. Equivalence point is when moles of acid = moles of base.

If the equivalence point is where acid and base are the same if you plug that into Hasselback equation Ph = pka + log (base/acid) you get pH= pka+ log (1/1) thus giving pH=pKa. This is the part that doesn't make sense to me. Since pH=pKa at half eq. point why is it you get the same with equivalence point? Like i know Since [A-]=[HA] at the half-eq point, the pH is equal to the pKa but that A- = HA sounds like your definition of the equivalence point!

 

[Rabolisk]

If the equivalence point is where acid and base are the same if you plug that into Hasselback equation Ph = pka + log (base/acid) you get pH= pka+ log (1/1) thus giving pH=pKa. This is the part that doesn't make sense to me. Since pH=pKa at half eq. point why is it you get the same with equivalence point? Like i know Since [A-]=[HA] at the half-eq point, the pH is equal to the pKa but that A- = HA sounds like your definition of the equivalence point!

That was (sort of) a mistake by ilovemcat. He doesn't mean the acid and base of the analyte (HA and A-), but rather, the initial amount of acid (HA) and the amount of the strong base added (OH-).

 

[flodhi1]

That was (sort of) a mistake by ilovemcat. He doesn't mean the acid and base of the analyte (HA and A-), but rather, the initial amount of acid (HA) and the amount of the strong base added (OH-).

AAAH! so equivalence point is the the number of moles of base added (NOT THE ANALYTE) equal to the original number of moles of acid present. and than the half equivalence point is just where the HA and A- equal each other in molar concentration?

 

[Rabolisk]

That's right.

Also, with respect to my first post, HH equation can also only be used when [H+] from water is negligible.

 

[Swagster]

This is seriously one of the most helpful threads I've read at SDN. Thank you everyone.

IloveMCAT: Where can I more examples like what you wrote?

 

[ilovemcat]

If the equivalence point is where acid and base are the same if you plug that into Hasselback equation Ph = pka + log (base/acid) you get pH= pka+ log (1/1) thus giving pH=pKa. This is the part that doesn't make sense to me. Since pH=pKa at half eq. point why is it you get the same with equivalence point? Like i know Since [A-]=[HA] at the half-eq point, the pH is equal to the pKa but that A- = HA sounds like your definition of the equivalence point!

HH equation (for a weak acid) is: pH = pKa + log (weak acid/conjugate base)

At half-half equivalence: [HA] = [A-]

The weak acid is at equilibrium prior to adding any strong base. Once strong base is added, the equilibrium is disrupted. The freely floating -OHs in the solution will deprotonate the weak acid (HA) into it's conjugate base (it loses its proton). As you add more and more strong base into the solution (of weak acid), more conjugate base is produced. Eventually you'll reach a point where you'll have equal amounts of HA and A-. This is the half-equivalence point (ie. buffer region).

At the equivalence point, there isn't anymore weak acid (HA). Instead there is 100% A-

If you continue to add more strong base (past the buffer region), you'll produce more A- (conjugate base ions). Infact, the more NaOH you add the more A- will be produced. Eventually the titration curve will shift upwards until you reach the equivalence point. The equivalence point occurs when 100% of the weak acid has completely been neutralized into it's conjugate base A-. At this point, there is no more weak acid to neutralize. This is the reason why we don't use the HH equation at the equivalence point.

 

[ilovemcat]

@Rabolisk - Thanks for the boost of confidence 😛 But right now, that's probably just wishful thinking though, haha.

@Swagster - No problem. If you have any questions, ask them in the Q&A thread. I'll try to help if I can, but Rabolisk is much more experienced in my opinion.

 

[SmurfinUSA]

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