pH=pka when 1/2 of HA is unreacted. WHY???

[Oh_Gee]

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from TBR:

"A buffer made by mixing 100 mL of 0.5MHOAc (Ka =1.8 x10^5) with 25 mL of
1.0M KOH has a pH approximately equal to which of the following values?
A. 0.2
B. 4.7
C. 7.0
D. 9.3
Solution
There are 0.05 moles of HOAc present and 0.025 moles of KOH present. This means that exactly half of an equivalent of strong base has been added to a weak acid, converting half of the original weak acid to its conjugate base. Half of the original weak acid remains unreacted. This means that pH = pKa. The value for
pKa is solved for as follows:
pKa =-log (1.8 x10"5) =-log 1.8 - log 10"5 =-log 1.8 - (-5) =-log 1.8 +5
Log 1.8 is less than 1, so the pH is greater than 4 and less than 5.
Choice B is best."


I did:
so given pH= pka + log [A-]/[HA]
log(1.8 x 10^5)=4.7
pH=4.7 +log (1/2)= 4.39

 

[jrhuynh94]

We can assume pH= pKa thus Ka = [H+]. Generally, when [H+] = 3 * 10^5, pH = exactly 4.5, anything less than 3 will bring the pH closer to 5 ( log(10^5) = 5 ), since 1.8 is less than 3, pH will be greater than 4.5 and less than 5 thus B is the best answer. Just remember, 3 * 10 ^n is the cut off.

 

[Oh_Gee]

We can assume pH= pKa thus Ka = [H+]. Generally, when [H+] = 3 * 10^5, pH = exactly 4.5, anything less than 3 will bring the pH closer to 5 ( log(10^5) = 5 ), since 1.8 is less than 3, pH will be greater than 4.5 and less than 5 thus B is the best answer. Just remember, 3 * 10 ^n is the cut off.

but is my math wrong? i'm confused about what you wrote

 

[jrhuynh94]

but is my math wrong? i'm confused about what you wrote

so given pH= pka + log [A-]/[HA]
log(1.8 x 10^5)=4.7
pH=4.7 +log (1/2)= 4.39

The definition at half equivalent is log[A-]/[HA] = 0, ratio of A-:HA is 1:1, therefore:
pH= pka + 0, pH=pKa
you don't add log(.5)

 

[popopopop]

The base is consumed by the weak acid, which is in excess, thus pH= pKa = left over acid with no base remaining.

 

[Oh_Gee]

so given pH= pka + log [A-]/[HA]
log(1.8 x 10^5)=4.7
pH=4.7 +log (1/2)= 4.39

The definition at half equivalent is log[A-]/[HA] = 0, ratio of A-:HA is 1:1, therefore:
pH= pka + 0, pH=pKa
you don't add log(.5)

but if you calculate out the mols then
pH=4.7 +log (.025/.05)=5

 

[jrhuynh94]

but if you calculate out the mols then
pH=4.7 +log (.025/.05)=5

A-/HA are conjugate base and its weak acid respectively, in this case they are OAc- and HOAc, these have the same [] at the half equivalent point. What you did was taking the log of [KOH]/[HOAc], which is not correct.

 

[Oh_Gee]

wait [A-] wouldn't be [OH-] would it?

 

[Oh_Gee]

wouldn't OAc- be .025 mols because .025 mols KOH reacts with .05 mols HOAc?

 

[jrhuynh94]

wait [A-] wouldn't be [OH-] would it?

No, A-is the conjugate base of HOAc which is OAc-.

 

[Oh_Gee]

what math did you do to get mols of OAc-?

 

[popopopop]

^.025 moles of OH- neutralizes .025 moles of HA (started with .05 moles) leaving you with 1:1 ratio of A-/HA. Log of that is 0, so pH= pKa + 0. Also, you wrote the Ka wrong. It's Ka = 1.8x10^-5. Using Berkeley's shortcut, you can say that pH = -(log2 + log10^-5) ---> -(.3 + -5) = -.3 - -5 = 4.7

Use the table to estimate logs, you can solve this in a few sconds if you remember that log 2 = .3.

 

[Oh_Gee]

ohhh wait.
so is the [HA] you use in the equation supposed to be the [HA] AFTER the base is added?

 

[popopopop]

^Yes lol, what do you think the strong base will do immediately after it hits the solution? For concepts, I like to reinforce with youtube/Chad's so a live person can explain it to me.

 

[Oh_Gee]

ohhh ok. one more question. the end of the Henderson equation uses concentrations of A- and HA but in the practice problems of TBR, they talk in terms of mols not []. why is that?

 

[popopopop]

It doesn't matter what you use since the ratio will end up being the same. Mols of A-/HA are floating in the same volume of solution.

 

[Oh_Gee]

thanks guys!

 

[BerkReviewTeach]

Great insights popop(etc...)!

 

[Czarcasm]

from TBR:

"A buffer made by mixing 100 mL of 0.5MHOAc (Ka =1.8 x10^5) with 25 mL of
1.0M KOH has a pH approximately equal to which of the following values?
A. 0.2
B. 4.7
C. 7.0
D. 9.3
Solution
There are 0.05 moles of HOAc present and 0.025 moles of KOH present. This means that exactly half of an equivalent of strong base has been added to a weak acid, converting half of the original weak acid to its conjugate base. Half of the original weak acid remains unreacted. This means that pH = pKa. The value for
pKa is solved for as follows:
pKa =-log (1.8 x10"5) =-log 1.8 - log 10"5 =-log 1.8 - (-5) =-log 1.8 +5
Log 1.8 is less than 1, so the pH is greater than 4 and less than 5.
Choice B is best."


I did:
so given pH= pka + log [A-]/[HA]
log(1.8 x 10^5)=4.7
pH=4.7 +log (1/2)= 4.39

Quick way to solve this: 100mL of 0.5M vs. 25 mL of 1.0M; I multiply the volumes x 10 to make this easier: so you have 0.5moles Acid and 0.25moles CB. From this, you should tell you have twice as much acid than CB. Had we had 10 times as much, the pH would be 1 pH unit lower than the pKa. But in this case, it's not even that much (only 2 times as much). The pH should change very slightly (less) than the pKa to favor the acidic side. From here, find the pKa which equals pH and from there, the value thats slightly less than the pKa is correct. 1.8x10^5 is between a pKa of 4 and 5, ...closer to 4.8. The only value a little less than this is choice B, 4.7.

 

[BerkReviewTeach]

The pH should change very slightly (less) than the pKa to favor the acidic side.

Be careful here. This is the same mistake that was being made that popopopop corrected. You are not plugging in values that have been added, you are plugging in values that result following reaction. The 0.05 moles of HOAc turn into 0.025 moles OAc- (after deprotonation) and 0.025 moles leftover HOAc. The result is a solution with equal parts HA and A-, so pH = pKa.

The thing you should notice is that OH- is a strong base that reacts and not the conjugate base of HOAc.

 

[Czarcasm]

Be careful here. This is the same mistake that was being made that popopopop corrected. You are not plugging in values that have been added, you are plugging in values that result following reaction. The 0.05 moles of HOAc turn into 0.025 moles OAc- (after deprotonation) and 0.025 moles leftover HOAc. The result is a solution with equal parts HA and A-, so pH = pKa.

The thing you should notice is that OH- is a strong base that reacts and not the conjugate base of HOAc.

Ah, glad you pointed that out. I totally overlooked that a strong base was added (and not the conjugate base like I assumed).