TBR acid/base

[cloak25]

How does propanoic acid exist after it has been added to water given that it has a pKa value of 5.0?

A. Fully deprotonated
B. 50% deprotonated
C. 25% deprotonated
D. Less than 1% deprotonated

answer: D

In this case, isn't the pH of water > pKa of the acid, so wouldn't it be fully deprotonated. 50% deprotontated occurs when pH = pKa. Why is it less than 1% deprotonated? Is it because the acid is a weak acid, so only dissociates partially?

---

Members don't see this ad.

 

[MedPR]

How does propanoic acid exist after it has been added to water given that it has a pKa value of 5.0?

A. Fully deprotonated
B. 50% deprotonated
C. 25% deprotonated
D. Less than 1% deprotonated

answer: D

In this case, isn't the pH of water > pKa of the acid, so wouldn't it be fully deprotonated. 50% deprotontated occurs when pH = pKa. Why is it less than 1% deprotonated? Is it because the acid is a weak acid, so only dissociates partially?

What's TBR's explanation? Amino acids are weak acids, yet they still fully protonate/deprotonate at their respective base/acid ends...

 

[cloak25]

Because the pKa for propanoic acid is 5.0, it is a weak acid, implying that it only partially dissociates in water. With a pKa value of 5.0, the Ka value is 1.0 x 10^-5. This means that for a 1.0M acid solution, the concentration of both H+ and conjugate base are 10^-2.5, which is less than 10^-2, or 1%. The dissociation is less than 1%. The Ka value is applied as follows:

Ka = [CH3CH2CO2-][H3O+]/[CH3CH2CO2H] = [CH3CH2CO2-]^2/[CH3CH2CO2H] = 1.0 x 10^-5 = 10 x 10^-6. This implies that the relative value of [CH3CH2CO2-] to [CH3CH2CO2H] is sqrt10 x 10^-3 to 1 which is roughly 3.1 x 10^-3 to 1. This value is less than one percent confirming choice D. Don't really get this part of the explanation.

 

[BerkReviewTeach]

Because the pKa for propanoic acid is 5.0, it is a weak acid, implying that it only partially dissociates in water. With a pKa value of 5.0, the Ka value is 1.0 x 10^-5. This means that for a 1.0M acid solution, the concentration of both H+ and conjugate base are 10^-2.5, which is less than 10^-2, or 1%. The dissociation is less than 1%. The Ka value is applied as follows:

Ka = [CH3CH2CO2-][H3O+]/[CH3CH2CO2H] = [CH3CH2CO2-]^2/[CH3CH2CO2H] = 1.0 x 10^-5 = 10 x 10^-6. This implies that the relative value of [CH3CH2CO2-] to [CH3CH2CO2H] is sqrt10 x 10^-3 to 1 which is roughly 3.1 x 10^-3 to 1. This value is less than one percent confirming choice D. Don't really get this part of the explanation.

That's basically the ice box setup.

HA <=> H+ + A-

Ka = xexp2/[HA]

Ka is 10exp-5, so if [HA] is in the normal area of 10-1 M for instance, then xexp2 = 10-6, making [H+] = 10-3 M. 10-3 M is about 1% of 10-1 M, so it's roughly 1%.

You can also figure that if it were fullydeprotonated, then pKa would be negative and if it were 50% deprotonated, then pKa would be around 0 (Ka = 1). To have a pKa as high as 5, you need it to barely dissociate.

 

[chiddler]

just use henderson hasselbalch equation.

 

[sillyjoe]

That's basically the ice box setup.

HA <=> H+ + A-

Ka = xexp2/[HA]

Ka is 10exp-5, so if [HA] is in the normal area of 10-1 M for instance, then xexp2 = 10-6, making [H+] = 10-3 M. 10-3 M is about 1% of 10-1 M, so it's roughly 1%.

You can also figure that if it were fullydeprotonated, then pKa would be negative and if it were 50% deprotonated, then pKa would be around 0 (Ka = 1). To have a pKa as high as 5, you need it to barely dissociate.

Why can't you use the Henderson Hassalbach equation for a problem like this? If you do, you get the wrong answer, so I am just curious as to why conceptually you can't use it.

 

[Teleologist]

Why can't you use the Henderson Hassalbach equation for a problem like this? If you do, you get the wrong answer, so I am just curious as to why conceptually you can't use it.

Of course you can use the HH equation if you knew the equilibrium concentration of the acid in advance, but if you knew both, that defeats the purpose of the problem.

just use henderson hasselbalch equation.

If I had a dollar everytime a student of mine pipes up "HH equation" to solve a pH problem then I wouldn't be a doctor. I'd drop outta school and buy me some virtual love, some physical love, and a mansion.

 

[sillyjoe]

Of course you can use the HH equation if you knew the equilibrium concentration of the acid in advance, but if you knew both, that defeats the purpose of the problem.



If I had a dollar everytime a student of mine pipes up "HH equation" to solve a pH problem then I wouldn't be a doctor. I'd drop outta school and buy me some virtual love, some physical love, and a mansion.

It sounded to me like someone earlier in the post was suggesting using the HH, but that would lead you to believe that the A- concentration is giant because you would mistakenly insert 7 = 5 + log (A-/HA)

Is it only when you know the final pH of the solution that you can use the HH to determine what the ratio is of A- to HA?

 

[Teleologist]

It sounded to me like someone earlier in the post was suggesting using the HH, but that would lead you to believe that the A- concentration is giant because you would mistakenly insert 7 = 5 + log (A-/HA)

Is it only when you know the final pH of the solution that you can use the HH to determine what the ratio is of A- to HA?

Buffer.

 

[sillyjoe]

Buffer.

Sorry what?