The following observations are made about a diprotic acid H2A:
(i...

Question

The following observations are made about a diprotic acid H2A:
(i) A 0.10 M solution of H2A has pH = 3.30.
(ii) A 0.10 M solution of the salt NaHA is acidic.
Which of the following could be the value of pKa2 for H2A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82?

Accepted Answer

hi everyone for this problem. It reads a 0.135 molar solution of di protic acid has a ph of 5.85. A 0.135 molar solution of the salt. K. H. A. Is basic. Which of the following is a reasonable value for the second P. K. A. Of this acid. Okay. So P. K. A. To this is what we're trying to solve and were given four answer choices here. Okay. So one thing we can start off with to solve this problem is look at what we're given. So we have both the malaria T and the ph of the dye protic acid. All right. So with these two pieces of information we can calculate the first P. K. A. Of the acid and we'll assume that the ph of the solution is only due to the first ionization of this acid. So let's go ahead and write out our reaction. Okay so we have our dia protic acid plus water is that equilibrium? And we have our hydro ni um ion and we have our acid. Okay so our K. A. Which is our first K. A. Is going to we're going to write out an expression for this. Okay. And that is our products over our reactant. Okay so we have our hydro ni um concentration times are acid over our react mints. And remember in our expression we don't include liquids or solids. So this is just going to be over I di protic acid. Okay and so from here we will use the ph to calculate the value of our hydro knee um ion concentration. And the way that we can calculate the P. H. Is by using the equation P. H. Is equal to the negative log of our hydro ni um ion concentration. So we have the P. H. And so we can go ahead and solve for our hydro knee. Um ion concentration. We know that our ph is 5.85. So it's going to be 5.85 is equal to the negative log of our hydro ni um ion concentration. So our hydro knee um ion concentration to rearrange this equation. This is what we're solving for. So it's going to be 10 Raised to the negative 5.85. That's how we're going to rearrange this equation. And so by rearranging this we can now solve for the right side and when we saw for the right side we get 1.41253 times 10 to the negative six molar. So this is our hydro knee um ion concentration which is also going to equal the concentration of our acid. Okay so these values are the same. So what that means is the concentration of our die protic acid is going to equal. We were given the starting mill arat. E so we have it's going to be the difference between these two values. So we started with 0.135 molar. And we're going to subtract our value that we just solved for our hydro ni um ion concentration. So that's minus 1. times 10 to the negative six Mohler. Okay so that's going to give us the value for our die protic acid. Okay so with these values we can now plug them into our K. A. Expression up at the top to solve for the value of K. A. One. Okay so let's go ahead and plug in. So we have our hydro knee um ion concentration times are acid concentration. Those values are the same. So we can just square it. So we have 1. times 10 to the negative six. Actually won't square. We'll write everything out completely. Okay so this is going to be times itself over our concentration of our die protic acid which we said is going to be the difference. Okay so we have 0. minus 1. times 10 to the negative six. So once we do this calculation and we plug it into our calculators we get let me write it in a different color K. A. One or our first K. A. Is equal to 1. 7797 times 10 to the negative 11. Okay so now that we know what this value is for our first K. A. We can convert this value of the K. A. To peek A. All right. And so that's what we're going to do. Now we're going to go from K A to P K. And the way we go from K A to P K A is P K A is equal to the negative log of K. A. Alright, so we have so we'll write P K. One. Okay? So we have P K A one is equal to the negative log of K. A. One. And we said that value was 1.47797 times to the negative 11. So P K A one Is equal to 10.8. So now we know our first peek a the question is asking us which of the following is a reasonable value for PKA. two of this acid. So a key thing we're going to need to know is for any weak acid I abbreviated as W A. The P K K two is going to be greater than the P K A one. So we know our P K A one is 10.8. Okay? And so we need to figure out what is our value for? R P K A two. So for any weak acid, P K two is going to be greater than P K one. So that means out of the answer choices we're given, we're going to need to find which P K A value is greater than 10.8. Okay? Because the one that's greater than 10.8 is going to be our weak acid. Okay. And it's going to be the one that's the reasonable value for that acid. So going back up to the top, we have A is 7.84. B is 8.25. C is 10.8 and D is 12.3. Okay, So A and B are both less than okay. They're both less than P K A two. Okay. And we said we're looking for the value that's greater than P K two and C is the same, is the same as P K one so see is not going to be correct. So the value that is greater than P K A One which is 10.8 is going to be d. So our correct answer is D 12.3. That's the end of this problem. I hope this was helpful.

The following observations are made about a diprotic acid H2A: (i) A 0.10 M solution of H2A has pH = 3.30. (ii) A 0.10 M solution of the salt NaHA is acidic. Which of the following could be the value of pKa2 for H2A: (i) 3.22, (ii) 5.30, (iii) 7.47, or (iv) 9.82?

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