Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Damiano0707 on March 16, 2018, 01:32:31 PM
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Hello everybody
I can't resolve this problem
You add 23.0mL of HCl 0.150M to a Na2SO3 solution. The final pH is 4.5. What are thr final moles of sulfite in solution?
A 0.00775
B 0.00694
C 0.00433
D 0.00345
I think a buffer solution is created but I can't say anything else.
Thanks to all
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Treating it like a buffer sounds like a good idea. Do you know the Henderson–Hasselbalch equation? Can you think how to apply it here?
(Actually plugging numbers into the dissociation constant definition is equivalently good approach; hardly surprising when you know where does the HH equation comes from).
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Yes I know it. The problem is that I don't know if the reaction is
Na2SO3+HCl--->NaCl+NaHSO3
Or
Na2SO3+2HCl--->2NaCl+H2SO3
In the first case I have SO32-/HSO32- and I could apply the equation. But there is still another problem. What are the moles of HCl that react?
In the second case I have SO32-/H2SO3 and I don't think I could apply the equation.
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What are pKa values of sulfuorus acid? What does the pH tell you - which pair of acid/conjugate base dominates the solution?
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What do you mean? Do you want to say in an acid base reaction weaker acids and bases are created? But HCl is a strong acid so either H2SO3 or NaHSOè can be created
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Wait a second... The pKa1 is much lower tanh pKa2 so SO32-/HSO3- dominates.
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I can't resolve the problem anyway
nb=ka2*na/[H+]=6.76*10^-6 It's not a result
I considered that the moles oh HCl are the same of NaHSO3.
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Okay, you need to make one important assumption when tackling this problem, which is you need to assume that HCl reacts with an excess of sodium sulfite solution.
If this is the case, what can we say about the equilbrium, and how will we set up HH equation?
EDIT: It looks like you made it above:
I considered that the moles oh HCl are the same of NaHSO3.
That's good, what happens next?
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As I said jI used the equation
nb = ka*na/[H+] = 6.76*10^-6
nb: number of moles of SO32-
ka: costant of equilibrium HSO3-=SO32-+H+
na: number of moles of HCl
[H+]: concentration given from pH
But the result is not among those one given in the problem. Why isn't correct?
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What are both pKa values, and please write both equilibrium expressions for them.
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Wait a second... The pKa1 is much lower tanh pKa2 so SO32-/HSO3- dominates.
No. Which one dominates depends on what is the pH. You would be right in the case of just dissociation, but here pH is forced on the system.
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The equilibriums are
HSO3-=SO32-+H+ pKa2=7.21
H2SO3=HSO3-+H+ pKa1=1.92
So I think the first equilibrium dominates because its pKa2 is nearer to pH but the calculus doesn't change and I don't get how to achieve the result
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[tex]K_a = \frac {[H^+][A^-]}{[HA]}[/tex]
rearranged:
[tex]\frac {[A^-]}{[HA]} = \frac {K_a}{[H^+]}[/tex]
You know everything on the right, try to use it to estimate relative amounts of H2SO3/HSO3-/SO32- - is there much H2SO3 present?
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[SO2-]/[HSO3-]=1.96*10^-3
But
[HSO3-]/[H2SO3]=ka1/[[H+]=379
[H2SO3] can be trascurated, but I don't get how I can continue. Do I have to consider that the moles of HCl are the same of HSO3-?
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Sigh, this question is a bit tricky, but actually much simpler than I thought (still, we got there thank's to the systematic approach).
Note, that from the numbers you have listed it is obvious that the solution contains mostly HSO3-. That means we are neither close to the pH=pKa1 nor pH=pKa2 (at which points there is a 50/50 ratio of the acid and conjugate base).
Do you know how to calculate pH of an amphiprotic salt solution?
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Actually no. Could you explain me?
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http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
Don't worry about the derivation, check out just the final formula - can you apply it here?
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Sorry but I don't think I can use it.
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You have listed both pKa values, why can't you use them?
(just in case: final formula is not the one listed on the page as the last)
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I understood the formula is
[H+]=√(ka1*ka2)
But the problem doesn't ask the final pH but the moles of SO32-.
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OK, try to apply it even if it doesn't do exactly what you need.
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ph=(pKa1+pKa2)/2= 4.565
It'similar to pH given
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And now?
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It'similar to pH given
Yes, that's exactly what I wanted you to spot!
Depending on the source of pKa values you will find that the pH given is either very close or just identical to the pH of the solution in question. Assuming the latter - what does it tell you about the composition of the solution?
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Thanks
I understood how to continue