Official DAT Destroyer Q&A Thread

[densaugeo]

Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!

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[DAT Destroyer]

I hope you can see that carbons 1, 4, 5, 8 are all equivalent......2, 3, 6, 7 are equivalent......so are 9 and 10. A Br can land at any of these positions to give 3 monobrominated products. No Br or any other atom is allowed at a ring junction.

I hope this helps.

Dr. Romano

 

[goprodent2]

I hope you can see that carbons 1, 4, 5, 8 are all equivalent......2, 3, 6, 7 are equivalent......so are 9 and 10. A Br can land at any of these positions to give 3 monobrominated products. No Br or any other atom is allowed at a ring junction.

I hope this helps.

Dr. Romano


View attachment 221175

Ok, I think just the double bond on left-most ring between C2and C3 just tripped me up a bit. I see what you mean now. Thank you, it helps a lot!

 

[Dallas Dentist]

For 2017 Destroyer, QR section.
On #96, I have never seen this formula for figuring out diameter. Any help on explaining?

 

[mrdeez]

Math destroyer 2016, practice test 2, #10.

"Donna said that in 6 years she will be 6 more than twice her daughter's age."

In the solution, the daughters age is represented as T (trish), and Donna's age is represented as D. The above quantity in quotations is represented as 2(T + 6) + 6. Doesn't the 2(T+6) quantity in this expression translate to "twice 6 more than her daughters age" instead of "6 more than twice her daughters age?"

If it was the former, as stated in the problem, then "6 more than twice her daughters age" would be represented as 2T + 6, where 2T is twice her daughter's age, and then you add six to that number. can someone explain to me how 2(T + 6) represents "6 more than twice her daughters age" and not "twice 6 more than her daughters age" Thanks!

 

[Counselingdentist]

Organic Chemistry Destroyer 2016 -- #259

Can anyone explain why the acyl chloride adds to the amine portion over being directed to the benzene ring? My brain is incredibly fried and I cannot seem to work it through.

 

[goprodent2]

2017 Destroyer, Gen chem #48

What is the Oxidation number for C5H12 in A. And for all the answer choices why, in the answer they only focus on one element. How do I know to find the oxidation number for oxygen in A, ZInc in B, potassium in C? Or do we go through all the oxidation numbers for each element?

 

[DAT Destroyer]

I have the 2017 version and the answer says choice D (area is quadrupled) is the right answer.

Day Destroyer QR question 17: the answer is correct. It says the correct answer is D

Math destroyer 2016, practice test 2, #10.

"Donna said that in 6 years she will be 6 more than twice her daughter's age."

In the solution, the daughters age is represented as T (trish), and Donna's age is represented as D. The above quantity in quotations is represented as 2(T + 6) + 6. Doesn't the 2(T+6) quantity in this expression translate to "twice 6 more than her daughters age" instead of "6 more than twice her daughters age?"

If it was the former, as stated in the problem, then "6 more than twice her daughters age" would be represented as 2T + 6, where 2T is twice her daughter's age, and then you add six to that number. can someone explain to me how 2(T + 6) represents "6 more than twice her daughters age" and not "twice 6 more than her daughters age" Thanks!

6 more than a quantity means 6 + the quantity. In 6 years her daughter will also grow by 6 years. So in 6 years her daughter will be (T + 6). Twice that is 2(T+6). 6 more than twice her daughter's age is then : 6 + 2(T+6)

Hope this helps.

 

[spdental]

Dr. Romano,

For question 318 O Chem 2017, I'm curious how you decide which alpha hydrogen to remove?

Additionally, questions 327 and 329 confused me, in 327 one of the answers chosen was C, however the charge on the carbon was wrong. I assumed this was a trick questions and that answer wasn't to be chosen because of the charge, same with question 329 for answers B and C.

 

[DAT Destroyer]

Dr. Romano,

For question 318 O Chem 2017, I'm curious how you decide which alpha hydrogen to remove?

Additionally, questions 327 and 329 confused me, in 327 one of the answers chosen was C, however the charge on the carbon was wrong. I assumed this was a trick questions and that answer wasn't to be chosen because of the charge, same with question 329 for answers B and C.

For 318, I removed the alpha hydrogen that was less sterically hindered as shown in the solution. For 327 the problem is fine. The question asked for the intermediate which can indeed be a charged specie. Shown is the enolate anion ,,,,note the two resonance forms. 329 is also fine as these charged structures represent intermediates we see in mechanisms such as halogenation, nitration, dehydration, and deamination. For more info, consult the Klein text to review intermediates in organic reaction mechanisms.

Hope this helps

Dr. Romano

 

[Wererew]

Hi Orgoman, in this video, can the nondisjunction of anaphase 2 also be n+1, n-1, n+1, n-1? And is there a nondisjunction where one cell gets all of it in animal cells? I believe that happens in plants but I haven't heard of it for animal cells. Like tetraploidy I think it was.

Same question... (haven't found an answer to this post)

Would n+1, n-1, n+1, n-1 be a possibility for failure at anaphse II? In that case how would we be able to distinguish between anaphase I or II?

 

[DAT Destroyer]

Same question... (haven't found an answer to this post)

Would n+1, n-1, n+1, n-1 be a possibility for failure at anaphse II? In that case how would we be able to distinguish between anaphase I or II?

View attachment 221239

This diagram sums it up best. Down syndrome is usually due to a Nondisjunction in Anaphase 1 or 2.....If in 1,,,,,we will get a Monosomy or a Trisomy....... If in Anaphase 2, we get Monosomy, Trisomy, or even a 50% to be normal. In about 5% of cases, a chromosome such as #14 adds on to #21 to produce what is called a Robertsonian result.

I hope this helps.

Dr. Romano

 

[Dallas Dentist]

Destroyer 2017, Biology #534

This question is asking about what a promoter site is. So I knew that D was correct in that the promoter site is the site on DNA at which RNA polymerase binds to initiate transcription. However, I am not understanding why B is also not correct?

 

[stu-dent1102]

Destroyer 2016 Question (QR section, problem 59)
Could someone please help me understand how to start the problem?
Initially, I plugged in l=6-w in order to try and find the variables, but that did not work.

The solution uses this property to answer the question: (a+b)(a-b)=a^2 - b^2
why would you use this property to solve this?

 

[DAT Destroyer]

2017 Destroyer, Gen chem #48

What is the Oxidation number for C5H12 in A. And for all the answer choices why, in the answer they only focus on one element. How do I know to find the oxidation number for oxygen in A, ZInc in B, potassium in C? Or do we go through all the oxidation numbers for each element?

I will show you a very easy way to do this. This DAT is not only about understanding the concepts, but SPEED and efficiency. It is easy to spot a redox reaction. Look for a metal standing all alone....without any charge. This will clearly indicate that B and C are redox reactions at rapid speed. Next. I teach my students another trick......Look for a diatomic molecule such as O2, N2, Cl2, etc......on one side of an equation. This is most likely a redox reaction. Choice A shows O2. on one side only......thus is also a redox.Choice D is not a redox,,,it is simply a double replacement reaction.

I hope this helps.

Dr. Romano

 

[goprodent2]

I will show you a very easy way to do this. This DAT is not only about understanding the concepts, but SPEED and efficiency. It is easy to spot a redox reaction. Look for a metal standing all alone....without any charge. This will clearly indicate that B and C are redox reactions at rapid speed. Next. I teach my students another trick......Look for a diatomic molecule such as O2, N2, Cl2, etc......on one side of an equation. This is most likely a redox reaction. Choice A shows O2. on one side only......thus is also a redox.Choice D is not a redox,,,it is simply a double replacement reaction.

I hope this helps.

Dr. Romano

Wow. Nicely put! I wish I had the magic cheat sheet of all these tips. Thank you Dr. Romano!

 

[deleted768895]

Hello Dr Romano,

I have a question about DAT Destroyer 2017 edition, Orgo Chem #37

The solution says that choices (b), (c) and (e) all have electron withdrawing groups. Are you referring to the benzene ring, or the additional groups? Furthermore, I notices someone else posted this question, but I couldn't actually find the answer, however, is the reason electron withdrawing groups decrease basicity because they allow resonance structures to be formed, which thus increases the stability, reduces basicity?
Thanks!

 

[DAT Destroyer]

Destroyer 2017, Biology #534

This question is asking about what a promoter site is. So I knew that D was correct in that the promoter site is the site on DNA at which RNA polymerase binds to initiate transcription. However, I am not understanding why B is also not correct?

Transcription start site, is where first RNA nucleotide binds. That is usually downstream (after the promoter site). Promoter site is used to guide RNA polymerase to the correct starting site, promoter is not used to initiate transcription.

Hope this helps.

 

[DAT Destroyer]

Hello Dr Romano,

I have a question about DAT Destroyer 2017 edition, Orgo Chem #37

The solution says that choices (b), (c) and (e) all have electron withdrawing groups. Are you referring to the benzene ring, or the additional groups? Furthermore, I notices someone else posted this question, but I couldn't actually find the answer, however, is the reason electron withdrawing groups decrease basicity because they allow resonance structures to be formed, which thus increases the stability, reduces basicity?
Thanks!

Yes.....I am referring to the electron withdrawing effect of the benzene ring. Aromatic amines are not as basic as aliphatic amines. The electrons are not localized. In B, we see a huge resonance effect, thus aniline is not that basic. In C, we have a neutral amide.These electrons are certainly not available for any sharing with a proton. In D, we have pyridine which is reasonably basic, BUT the electrons are held rather tightly in an sp2 orbital. In E, we have some resonance along with an electron withdrawal effect from the NO2 group. Only choice A shows an amine with electrons that are not only localized but quite available for electron donation. This molecule namely cyclopentanamine in Choice A can be a good base. If electrons are LOCALIZED, we have an amine that can function as a base. If these electrons are delocalized by either induction, or resonance we lose much of the basicity.

I hope this helps.

Dr. Romano

 

[DAT Destroyer]

Destroyer 2016 Question (QR section, problem 59)
Could someone please help me understand how to start the problem?
Initially, I plugged in l=6-w in order to try and find the variables, but that did not work.

The solution uses this property to answer the question: (a+b)(a-b)=a^2 - b^2
why would you use this property to solve this?

The answer choices given represent the length and the width. Try each one to see which one will give: L*W=7 and 2(L+W)=12
You don't have to use the property given, you can simply foil and you get the same answer. It's faster if you use the property

Hope this helps, if it did please like my post.

Thanks..

 

[DAT Destroyer]

MATH DESTROYER Test 3 #10... How are we getting 70 degrees? This is my diagram so far. I'm so stuck.

Since arc BC subtends an angle of 40° , arc AB will subtend an angle of 140° since AC is the diameter of the circle. Now by definition the measure of angle BCA will be half the measure of the arc AB.

If this helps like our answer..

Thanks...

 

[DAT Destroyer]

For 2017 Dat destroyer, when looking at the answers, why is is the formula for 27 different than 20. why is there the extra (r!) in the denominator? I don't know when to use/incorporate the extra (r!) in the formula.

Thanks for the help!

For number 20, the order in which the horses finish a race matters, hence it's a permutation. For #27 we are only looking to select 3 out of 12. The order doesn't matter, hence it's a combination.

Hope this helps..

 

[goprodent2]

2017 DAT destroyer Gen chem

For #37, we weren't able to determine the rate law bc not enough info. But for #70 we are able to. Can you explain when can we and when we can't determine a rate law.

Also for #70, the rate law has the concentrations of the final rxn with the coefficients being the exponents in the rate law. I assume this was a coincidence as this won't able be true for all rxn's, but the the rate law will always be the concentrations of the final rxn? Thank you for your help.

Edit: one more question for #71 D, why does Sulfur form 6 bonds. I thought it can only form 4? I know boron can form 3 bonds and be 'happy' as in stability. Bc if one of the oxygens formed another bond with Sulfur then, Sulfur would've have 4 bonds, full octet. and all electrons accounted for.

 

[RDHtoDDS_40]

For 2017 DAT Destroyer Organic Chemistry:
Q: #19: Consider the following compound molecular formula C20H44N2O. Which formula cannot be present?
A- amide

In the answer key, why is N subtracted from the molecular formula and then compared to the "nearest alkane?"

 

[RDHtoDDS_40]

For 2017 DAT Destroyer Organic Chemistry:
Q: #19: Consider the following compound molecular formula C20H44N2O. Which formula cannot be present?
A- amide

In the answer key, why is N subtracted from the molecular formula and then compared to the "nearest alkane?"

@orgoman22

 

[yoh95]

Is it helpful if we point out typos in the Destroyers?

Biology #138 answer choices are written a,b,a,b,c in the 2017 version

Edit: Biology section #342 Answer choice says its C but it should be E (from reading the explanation)

 

[deleted768895]

Hello Dr Romano,

Im very confused about a general chemistry concept, that has just come up on the DAT Destroyer 2017 edition #34. The question asks which one of the ions has the largest radius. These appear to be an isoelectronic series, as they all have the same number of electrons. I had the understanding that in an isoelectronic series, since electron # is the same, the trends in atomic radius do not play a part, more so, whichever ion has the largest atomic number (and thus # of proton), will have the greatest nuclear effective charge, greatest 'pull' into the centre, and subsequently, will be the smallest ion. Conversely, the largest radius will be the ion with the smallest atomic number. However, the correct answer choice is S-- which does not have the smallest atomic #, and the solution says that the greater the negative charge, the more electrons have been gained, and thus the radius has increased in size. This seems to go against what I have learnt about isoelectronic series?

Please could you help clarify this for me.
Thanks!

 

[Dallas Dentist]

Just a general question regarding polarity, but questions 336 and 374 of Gen Chem 2017 could be referred.

#374: So from what I've looked up XeFe4 is considered non polar because all of the F atoms will cancel dipole moments and then the two lone pairs repel each other.
So then in #336, How come SBr2 is considered a polar molecule if it has two bromines whose dipoles will cancel out and also two lone pairs?

 

[DAT Destroyer]

For 2017 DAT Destroyer Organic Chemistry:
Q: #19: Consider the following compound molecular formula C20H44N2O. Which formula cannot be present?
A- amide

In the answer key, why is N subtracted from the molecular formula and then compared to the "nearest alkane?"

For problem #19......I found what is called the degree of unsaturation. The solution shows how to do it. If the degree of unsaturation is zero.....it means that the compound has no rings, double or triple bonds. Thus we get a zero,,,,,hence the compound CANNOT be an amide,,,,,since an amide contains a double bond.

Hope this helps.

Dr. Romano

 

[DAT Destroyer]

Hello Dr Romano,

Im very confused about a general chemistry concept, that has just come up on the DAT Destroyer 2017 edition #34. The question asks which one of the ions has the largest radius. These appear to be an isoelectronic series, as they all have the same number of electrons. I had the understanding that in an isoelectronic series, since electron # is the same, the trends in atomic radius do not play a part, more so, whichever ion has the largest atomic number (and thus # of proton), will have the greatest nuclear effective charge, greatest 'pull' into the centre, and subsequently, will be the smallest ion. Conversely, the largest radius will be the ion with the smallest atomic number. However, the correct answer choice is S-- which does not have the smallest atomic #, and the solution says that the greater the negative charge, the more electrons have been gained, and thus the radius has increased in size. This seems to go against what I have learnt about isoelectronic series?

Please could you help clarify this for me.
Thanks!

Only S--, K+ and Ca++ are isoelectronic. If you calculated the effective nuclear charge on S-- we get +6.....K+ is + 9...and Ca ++ is 10.........This alone supports the fact that S is largest. The literature values show that S-- is 184pm, K+ is 133pm, and Ca++ is 99.

I hope this helps.

Dr. Romano

 

[spdental]

For Gen Chem 2017 Question 21 - I understand that increasing molecular mass increases the van der waal forces which in turn will make an element a gas or liquid at standard temp, is there a certain point where a gas turns into a liquid? I was confused by how I would figure this out, unless of course this is just a memorization problem.

Thanks!

 

[goprodent2]

Im doing DAT destroyer QR, #73

and im wondering, is there a difference between cos^2 (x) vs. cos (x)^2. If there is a difference, can you please explain how that might differ? Because in the 'formula sheet' beginning of QR, it shows cos^2 (x) but in the answer for 73, it re-writes it the other way. Thank you for the help.

 

[deleted768895]

Only S--, K+ and Ca++ are isoelectronic. If you calculated the effective nuclear charge on S-- we get +6.....K+ is + 9...and Ca ++ is 10.........This alone supports the fact that S is largest. The literature values show that S-- is 184pm, K+ is 133pm, and Ca++ is 99.

I hope this helps.

Dr. Romano

How did you calculate the effective nuclear charge above? Thanks!

 

[deleted768895]

Hello Dr Romano,

I have a question on the DAT Destroyer 2017 edition, General Chemistry #50.

Although I agree, that both acids and their corresponding salt are in a 1:1 ratio, thus according to the Henderson Hasselbach, pH = pKa + Log (1) --- which is 0... Thus the equation becomes pH = pKa. However, without having the Ka or pKa values of the acids, how can we conclude that their pH values will be the same?

 

[DAT Destroyer]

For Gen Chem 2017 Question 21 - I understand that increasing molecular mass increases the van der waal forces which in turn will make an element a gas or liquid at standard temp, is there a certain point where a gas turns into a liquid? I was confused by how I would figure this out, unless of course this is just a memorization problem.

Thanks!

This is a very basic problem . Do NOT overthink this. There are 118 elements on the table.....2 are liquids....Hg and Br2. Gases include H2, O2, N2, Cl2, F2, and group 18. All the rest are SOLIDS !!!!! This is something that is easy to memorize, and must be with you FOREVER !!!!! Keep hammering away.

I hope this helps.

Dr. Romano

 

[deleted768895]

Hi Dr Romano,

Sorry for the numerous questions, but I try to google answers and can't find a correct response. For question #76 on the DAT Destroyer 2017 edition, Organic Chemistry, Why is the Nitrogen Sp2 hybridized? So its got 3 bonding connections and a lone pair, so I would've thought it'd be a SP3? It thus seems the lone pair of electrons are not being included, why is that?

 

[Wererew]

Hi Dr Romano,

Sorry for the numerous questions, but I try to google answers and can't find a correct response. For question #76 on the DAT Destroyer 2017 edition, Organic Chemistry, Why is the Nitrogen Sp2 hybridized? So its got 3 bonding connections and a lone pair, so I would've thought it'd be a SP3? It thus seems the lone pair of electrons are not being included, why is that?

Hello,

The lone pair is participating in the aromatic system (to satisfy the 4n+2 electrons) so it is delocalized into a p orbital. This would make the Nitrogen atom sp2 hybridized. I believe there are no sp3 hybridized orbitals in an aromatic ring due to resonance...

The other Nitrogen in the ring is also sp2 hybridized but the lone pair is localized into a separate sp2 orbital... (not participating in the aromatic system, hence perpendicular to the p orbitals)

The nitrogen on the amine group is sp3 hybridized

 

[Wererew]

How did you calculate the effective nuclear charge above? Thanks!

I don't think you need to know how to calculate these values..

For the isoelectronic series, remember that anions have a bigger radius than cations and vise versa.

 

[goprodent2]

Hello,

The lone pair is participating in the aromatic system (to satisfy the 4n+2 electrons) so it is delocalized into a p orbital. This would make the Nitrogen atom sp2 hybridized. I believe there are no sp3 hybridized orbitals in an aromatic ring due to resonance...

The other Nitrogen in the ring is also sp2 hybridized but the lone pair is localized into a separate sp2 orbital... (not participating in the aromatic system, hence perpendicular to the p orbitals)

The nitrogen on the amine group is sp3 hybridized

I have a question, so for the other nitrogen with lone pairs not participating in the ring, you say it's sp2. would it always be the same hybridization as the nitrogen if the lone pairs are not participating in the ring? And if they were participating in the ring, they will always be in p orbital as the destroyer says? thanks for your help.

 

[Wererew]

I have a question, so for the other nitrogen with lone pairs not participating in the ring, you say it's sp2. would it always be the same hybridization as the nitrogen if the lone pairs are not participating in the ring? And if they were participating in the ring, they will always be in p orbital as the destroyer says? thanks for your help.

Yes, I believe that is the case... if you look at the compound furan, one of the lone pair of the oxygen is delocalized into a p orbital to be aromatic and the other is localized into an sp2 orbital not participating

 

[goprodent2]

Can someone please explain what problem #84 QR, 2017 DAT destroyer says.

It Recall .75 is 3/4 raised to the sqrt(3) ?? is this a typo or can you explain how that works.

Thank you.

 

[goprodent2]

Also, DAT destroyer #85,86,89

Permutation vs. combination.

For 85 and 86 the formula used was permutation, Dr.Romano, can you explain how this asks for an ordered sequence. and when would it be a combination.

For 89, the reason this is permutation (ordered) is because the letter cannot be used more than once, or what is the key to know if this is ordered or non-ordered.

Thank you for your help Dr. Romano!

 

[DAT Destroyer]

Can someone please explain what problem #84 QR, 2017 DAT destroyer says.

It Recall .75 is 3/4 raised to the sqrt(3) ?? is this a typo or can you explain how that works.

Thank you.

0.75 can be written as 3/4 and 27/64 can be written as (3/4)^3.
(3/4)^y = (3/4)^3. Because the bases are the same we have y = 3

 

[goprodent2]

0.75 can be written as 3/4 and 27/64 can be written as (3/4)^3.
(3/4)^y = (3/4)^3. Because the bases are the same we have y = 3

Ok, I inserted the photo of it below. So, when you wrote (3/4) being raised to Sqrt(3), that meant that 3/4 will be cubed? Thank you for the clarification Dr. Romano

 

[spdental]

Dr Romano,

I have a question on 2017 O Chem 35. This doesn't have to do directly with the question, however it'd be great to know. For option C vs option D, if we were to be asked which is a strong ACID would the choice be D? Because there would be a greater steric repulsion from the tert group?

 

[goprodent2]

Dr Romano,

I have a question on 2017 O Chem 35. This doesn't have to do directly with the question, however it'd be great to know. For option C vs option D, if we were to be asked which is a strong ACID would the choice be D? Because there would be a greater steric repulsion from the tert group?

I literally just watched a video on this. If you look up CARDIO organic chem on google, (link: 5 Key Factors That Affect Acidity in Organic Chemistry — Master Organic Chemistry) this is what Mike's videos (on BC) talks about. He compares how to figure out which is more acidic. If you look it up, im sure it will help you out. It's ch2 of Mikes Orgo videos if you have Bootcamp.

Hope this helps!

 

[spdental]

I literally just watched a video on this. If you look up CARDIO organic chem on google, (link: 5 Key Factors That Affect Acidity in Organic Chemistry — Master Organic Chemistry) this is what Mike's videos (on BC) talks about. He compares how to figure out which is more acidic. If you look it up, im sure it will help you out. It's ch2 of Mikes Orgo videos if you have Bootcamp.

Hope this helps!

I understand the inductive effects, just in this case it's a very close molecule so I was wondering if a methyl group moved one over would change that.

 

[goprodent2]

I understand the inductive effects, just in this case it's a very close molecule so I was wondering if a methyl group moved one over would change that.

uhm, i really dont know. Im stuck learning orgo over again, just thought that link would help out.

 

[deleted768895]

Hello,

The lone pair is participating in the aromatic system (to satisfy the 4n+2 electrons) so it is delocalized into a p orbital. This would make the Nitrogen atom sp2 hybridized. I believe there are no sp3 hybridized orbitals in an aromatic ring due to resonance...

The other Nitrogen in the ring is also sp2 hybridized but the lone pair is localized into a separate sp2 orbital... (not participating in the aromatic system, hence perpendicular to the p orbitals)

The nitrogen on the amine group is sp3 hybridized

Thank you!! So how would you know which nitrogens lone pair are participating in the aromaticity? Because in this example, only one lone pair from one nitrogen was necessary to participate in the aromatic system to achieve the 4n + 2. But how do we know it was that nitrogen (the one of the left) as compared to the other one?

 

[Florida Dent]

2017 DAT Destroyer
Biology Question #297

Is the answer 4N during anaphase because a somatic cell is already 2N (46 chromosomes) so when the sister chromosomes split and the chromosome # is now 96 you do 2N x 2? I answered 2N because the chromosome number of somatic cells is 46 and during anaphase it is 96. I am assuming I was supposed to consider that they were already 2N but I would like to make sure.

 

[deleted768895]

Hello Dr Romano,

I have a question on the DAT Destroyer 2017 edition, General Chemistry #50.

Although I agree, that both acids and their corresponding salt are in a 1:1 ratio, thus according to the Henderson Hasselbach, pH = pKa + Log (1) --- which is 0... Thus the equation becomes pH = pKa. However, without having the Ka or pKa values of the acids, how can we conclude that their pH values will be the same, if pH = pKa?