Topic: Turn case solutions
June 20, 2019 / By Debbie Question:
The following solutions contain a weak acid HA (pKa = 5) and its sodium salt, NaA
Solution (a) contains: 3 HA and 5 A-
Solution (b) contains: 4 HA and 4 A-
Solution (c) contains: 4 HA and 5 A-
Solution (d) contains: 6 HA and 4 A-
Question 1: Which solution has the lowest pH?
Answer: solution d because it has the most HA?
Question 2: Which has the highest pH?
Answer: solution a because it has the least HA?
Question 3: How many different species (not including water and Na+ ion) are present after the addition of two H+ ions to this solution a?
Question 4: How many different species (not including water and Na+ ions) are present after the addition of two OH- ions to solution b?
Answer: Not sure
Thanks for taking the time to look over it.
Britannia | 9 days ago
NO!! WRONG! Not even close! **
** At least as I interpret the problem. It is extremely poor chemistry to claim that a solution contains 3 HA. WTF does that mean???!!! IDK. I am assuming it means that [HA] = 3 , specifically 3 Molar, which is the units of [....] last I heard. It is even worse when someone idiotically claims that two H⁺ ions are added. Slap your teacher up on side of the head, hard. Two ions would change the concentration insignificantly (unless, of course you are talking about attoliters). Almost criminal incompetence. So, ask your teacher YTF s/he thinks phrasing the problem that way is even acceptable. It is not.
pKa is by definition -log(Ka)
Ka is by definition (well, for beginner students)
for the HA ⇌ H⁺ + A⁻ equilibrium, Ka = [H⁺][A⁻] ÷ [HA]
You apparently think you can do it in your head. Good luck with that.
You are given four different values for [A⁻] and [HA]
you solve the equation for [H⁺], it is trivial arithmetic.
In the case you are given, it turns out that answer 1 is d and 2 is a, but NOT for the reason you state!!
For instance if d were 3 HA and 1 A⁻ it has the LEAST HA and yet is still lowest pH.
Try it. And never do it the lazy way again.
Q3 In solution a there is already HA, A⁻ and H⁺ (actually H₃O⁺ and H(H₂O) ͓⁺ ions, a bare proton is virtually never found, H₃O⁺ predominates). There is also (as you should know) some OH⁻ "floating around". Not much, but some. If two moles of H⁺ are added, then you could ignore the OH⁻, with confidence. So, I count 3 at least and if we wanted to stretch a point 4.
Q4 Again assuming your teacher is an idiot, and that "two OH⁻ ions" means two moles of OH⁻, then
you should assume all of it reacts with HA to form H₂O and A⁻ but you have 4 moles of HA, so that leaves 2 HA and 6 A⁻. And doesn't change the count, you still have HA, H⁺, A⁻ and a bit more OH⁻
But plugging into the equation to determine pH, you get a pretty acidic mix, so again the OH⁻ can be ignored.
Note that in neither Q3 nor Q4 was the counter-ion identified. You can't add just H⁺, you can't add just OH⁻, both need a counter-ion. If you used, say HCl as the source of H⁺, the Cl⁻ ion would be added to the count. If you used, say KOH as the source of OH⁻, then the K⁺ ion would add to the count. You could add NaOH in Q4 and ignore the Na⁺, but there is nothing you can add that makes any sense in Q3 which would not change the count. Why couldn't you add HA as the source of H⁺ ?
You could but that would increase not just the [H⁺], but both the [HA] and the [A⁻], too.
While I have no idea why, the answer to both Q3 and Q4 is 2. (I had this same one for my assignment.)
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