IMAT 2021 51 ACID BASE key concepts

Answer: A
correct me if im wrong:
Reasoning for answer is,

  1. wrong

  2. Wrong. Both nitric and sulfuric acid are strong acids and will dissociate fully in into ions.

  3. True.

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for 3 what do u mean by “if it was a monoprotic strong acid, pH of 2 would be correct”? wouldn’t the pH if it were monoprotic simply be -log[0.05] which is around 1.3? also for diprotic we calculate ph by multiplying the concentration by 2 as in -log[(0.05x2)], right?

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1- Sulfuric acid is a strong acid which dissociates completely in water. All strong acids protons have a pKa-value smaller than 4. In this case sulfuric acid has -3.0 and 1.99 as pKa1 and pKa2 (according to my uni data).

2- Nitric acid has a pKa of -1.4 and is fully dissociated in water. So, you are right.

3- You have in your #1 statement said, H2SO4 does not release all its protons, and now you say in #3 that it releases more than one hydrogen ion!
We have monoprotic acids which are not strong (having pKa over 4).
So, we need to calculate [H+] to answer the question. In this case pH will be 1.


Thank you for your feedback. There are numerous information out there saying that a solution of H2SO4 will not completely release 2 protons at room temp and there are specific calculation methods for that but for IMAT level, I guess the approximate pH would be 1.

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Thank you for your reaction. Though I do not agree with dissociation issue.
Calculations methods is another issue. In this question the ph will be precisely 1, since pH=-log[H+]=-log[2 . 0.05]=1. Cheers. :pray:


the best way to solve this is by using log rules. so we know that pH = -log[H+], and as the first proton of H2SO4 fully dissociates we know that the ph = -log[0.05]. Now, we need to simplify 0.05 into the standard form which will give us pH = -log[5 * 10^-2 ]. This can be further simplified to:
pH = -(log[10^-2] + log[5])
pH =-log[10^-2] -log[5]
now we know that log of 10^-2 = -2
pH= -(-2) -log[5]
pH = 2 - log[5]
You should know that the log (to the base 10) of any number less than 10 will be less than 1, and any number more than 1 will be more than 0. Therefore we know that log[5] equals a number between 0 and 1. This will mean that Y’s pH, regardless of whether the second proton dissociates partially or not, will be less than 2.

I hope this helps.