Hey könnte jemand mir bei dieser Chemie Aufgabe helfen?

CfragtFragenByCfragtFragen

Der pH-Wert einer wässrigen Natriumacetat-Lösung mit einer Konzentration von 4 mol L-1 beträgt

9,68 (pKg = 9,25). Zu einem Liter dieser Lösung werden 98 g reine (100 %-ige) Schwefelsäure

hinzugegeben. Wie groß ist der pH-Wert der entstehenden Lösung?

Atommassen: S: 32 u; H: 1 u; O: 16 u

(2 votes)
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indiachinacook
indiachinacook
1 year ago

So we have four moles of sodium acetate and one mole of sulfuric acid in the same solution. Each molecule H2SO4 eats two acetates:

H2SO4 + 2 CH3CO2 ̄ SO42 ̄ + 2 CH3COOH

and forms therefrom two molecules of acetic acid.

So we converted from the initially four moles of acetate two to acetic acid, and two remain. Finally, we get a solution with the same amount of acetate as acetic acid, i.e. a symmetrical buffer, i.e. pH=pKa=4.75.

A few questions could still be raised:

  • Why is the pH of the sodium acetate solution there? I don’t know.
  • The base constant of the acetate is given, which we ourselves have to convert into the acid constant (14-).
  • How do we know that the sulfuric acid releases both protons and not just one (→hydrogensulfate HSO4 ̄)? The HSO4 ̄ is a medium-strength acid (pK2≅2), so it is completely deprotonated at pH “2”.
  • Why are the solutions so highly concentrated? I don’t know. Since the equations no longer function well without activity coefficients.
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CfragtFragen
Author
CfragtFragen
1 year ago
Reply to  indiachinacook

Oh, no, this is the pKa of sodium acetate. But thank you! I just don’t understand where you know the pKa is 4.75?

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indiachinacook
indiachinacook
1 year ago
Reply to  CfragtFragen

14-9.25=4.75

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Baumbaumast
Baumbaumast
1 year ago
Reply to  indiachinacook

Hey, I don’t understand how to adjust the reaction to 2 CH3COOH, so why the two in front? Without 2 the equation is stoichiometrically balanced? Thank you

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indiachinacook
indiachinacook
1 year ago
Reply to  Baumbaumast

The reaction equation had a typing error, right is it

H2SO4 + 2 CH3CO2 ̄ SO42 ̄ + 2 CH3COOH

Somehow, the two before the acetate were lost when tipping, when I considered it correctly (each H2SO4 eats) two Acetates’. Of course I better get it in the answer.

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indiachinacook
indiachinacook
1 year ago

I briefly indicated this in the answer: The sulfuric acid is two-basic, so it can theoretically react with the acetate in two ways (it can release one or two H+):

H2SO4 + CH3CO2 ̄ ⟶ HSO4 ̄ + CH3COOH

H2SO4 + 2 CH3CO2 ̄ SO42 ̄ + 2 CH3COOH

Both reactions could in principle proceed and do so under different conditions:

  • HSO4 ̄ is stable only at pH22; at a higher pH there is its H+. This is seen in that the second pKa of the H2 SO4 (i.e. exactly the protolysis of HSO4 ̄) has a value of ≅2.
  • Conversely, SO42 ̄ at pH22 is stable, from more or less the same greens, only seen in reverse.

When solving your task, you do not know the pH of the solution at the end, so you must appreciate: if you expect a pH>2, then you select the 1:2 ratio between H2SO4 and CH3CO2 ̄, otherwise 1:1. As I have experience with such things, I have made it right, but if I had found myself for the matter, then I would have noticed it at the end because the result does not fit to accept pH2. Here false Invoice:

H2SO4 + CH3CO2 ̄ ⟶ HSO4 ̄ + CH3COOH

Each molecule H2SO4 eats an acetate and converts it to acetic acid; three moles of acetate remain. We now have an asymmetrical acetic acid/acetate buffer, and we continue with Henderson–Hasselbalch

pH = pKa + lg (acetate/acetic acid) = 4.75 + lg(3⁄1) = 5.22

And this result cannot be correct because at pH≅5 no HSO4 ̄ can float around in the solution.

If we modify the task in such a way that one liter of 2-molar H2 SO4 is reacted with 1 mol of solid CH3 CO2, then one mole of acetic acid +HSO4 ̄ plus one mole of unused H2 SO4, and the solution has quite exactly pH=2.

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Baumbaumast
Baumbaumast
1 year ago

Jaa understands achso thank you, but how do you know that? So how do I know, then sulfuric acid the two acetate, ‘freeze’, you can see that on the basis of the equation of reaction, but can you find that otherwise? Evtl. Because they thought h2so4 was 2 protonig, so? I’m trying to understand the task, but it’s a bit difficult for me! Thank you

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Luzifer234
Luzifer234
11 months ago

In order to calculate the pH of the resulting solution, we first have to take into account the pH of the sodium acetate solution and then the action of the added sulfuric acid.

1. Calculate the pH of the sodium acetate solution:

We know that the pKb value for sodium acetate is 9.25.

Since the pH is given (9.68), the solution is basic. We can calculate the pOH value: pOH = 14 – pH = 14 – 9.68 = 4.32.

Then we use the equation pOH = pKb + log([Acetat]/[Essig Acid]) (for a weak base) to calculate [Acetat]/[Essig Acid].

9,25 = -log(Kb) + log([Acetat]/[acetic acid])

Since [acetate] and [acetic acid] are present in the same concentration and are therefore shortened, we can simplify:

9,25 = -log(Kb)

Kb = 10^(-9,25)

Since we know Kb, we can now calculate [Acetat]/[acetic acid].

Two. Calculate the pH after addition of sulfuric acid:

The added sulfuric acid is completely dissociated in water and added to the solution H3 O+.

The concentration of H3 O+ is determined by the dissociation of sulfuric acid:

H2SO4 -> 2H+ + SO4^2-

The concentration of H3 O+ is therefore 2 * (98 g/98 g/mol) mol/L.

Then use the equation pH = -log([H3O+]) to calculate the pH value.

I hope that will help you solve the task! If you have more questions, let me know.

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