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# Hard Acids and Buffers Q watch

1. I have partially filled the table in; the bold numbers were already given.

My questions:
1) Is it "double equivalence" because Ba(OH)2 is diprotic? What does it mean?
2) Is there a way of working out half equivalence point of W without using pH = pKa? I mean, we know the volume and concentration and pH of the weak acid and we know the volume and concentration of Ba(OH)2.
3) The double equivalence for S doesn't make sense. I have 0.00144 moles of barium hydroxide being added to 0.00288 moles of S. This means because of the 1:2 ratio (Ba(OH)2Cl), 0.00144 x 2 moles of HCl will react leaving 0.00288 moles - 0.00288 moles of H+ in the solution and 0 moles of OH- in the solution :s
4) I'm also finding it difficult to understand the context. We have a base which undergoes 2 seperate reactions, right? Then are both of these reactions also in 1:2 ratio? Here's what I mean:
Ba(OH)2 + HCl -> etc.
Ba(something) + HCl -> etc
Only when adding the 2 equations we get the 1:2 ratio, individual reactions are 1:1

Really confused by this Q guys, would really appreciate some help
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2. (Original post by ps1265A)
I have partially filled the table in; the bold numbers were already given.

My questions:
1) Is it "double equivalence" because Ba(OH)2 is diprotic? What does it mean?
2) Is there a way of working out half equivalence point of W without using pH = pKa? I mean, we know the volume and concentration and pH of the weak acid and we know the volume and concentration of Ba(OH)2.
3) The double equivalence for S doesn't make sense. I have 0.00144 moles of barium hydroxide being added to 0.00288 moles of S. This means because of the 1:2 ratio (Ba(OH)2Cl), 0.00144 x 2 moles of HCl will react leaving 0.00288 moles - 0.00288 moles of H+ in the solution and 0 moles of OH- in the solution :s
4) I'm also finding it difficult to understand the context. We have a base which undergoes 2 seperate reactions, right? Then are both of these reactions also in 1:2 ratio? Here's what I mean:
Ba(OH)2 + HCl -> etc.
Ba(something) + HCl -> etc
Only when adding the 2 equations we get the 1:2 ratio, individual reactions are 1:1

Really confused by this Q guys, would really appreciate some help

Funny old question.

18ml of 0.16 M HCl is used = 0.00288 mol
This would need 0.00288/(2 x 0.12) litres of Ba(OH)2 = 12 ml for equivalence

The 'double equivalence' is when the moles of Ba(OH)2 equal the moles of HCl initially. (I've never heard of this or used it before, I suppose it's a way of calculating the moles of HCl with an easier calculation??)

The half-equivalence point will be the same for the weak acid HX as for HCl as they have the same molarity and are reacting with the same base.
3. (Original post by charco)
Funny old question.

18ml of 0.16 M HCl is used = 0.00288 mol
This would need 0.00288/(2 x 0.12) litres of Ba(OH)2 = 12 ml for equivalence

The 'double equivalence' is when the moles of Ba(OH)2 equal the moles of HCl initially. (I've never heard of this or used it before, I suppose it's a way of calculating the moles of HCl with an easier calculation??)

The half-equivalence point will be the same for the weak acid HX as for HCl as they have the same molarity and are reacting with the same base.
18ml of 0.16 M HCl is used = 0.00288 mol
This would need 0.00288/(2 x 0.12) litres of Ba(OH)2 = 12 ml for equivalence

So HCl: Ba(OH)2, you've done c1v1=2(c2v2), but why 2 when there's twice as much HCl than Ba(OH)2, it should be other way around?
4. (Original post by ps1265A)
18ml of 0.16 M HCl is used = 0.00288 mol
This would need 0.00288/(2 x 0.12) litres of Ba(OH)2 = 12 ml for equivalence

So HCl: Ba(OH)2, you've done c1v1=2(c2v2), but why 2 when there's twice as much HCl than Ba(OH)2, it should be other way around?
In terms of moles there is twice as much HCl needed as barium hydroxide..
5. (Original post by charco)
In terms of moles there is twice as much HCl needed as barium hydroxide..
I still can't get my head around it... Based on what you've said, there twice as many moles of HCl than Ba(OH)2, but you need twice as much Ba(OH)2 to neutralize this 2 mols?
6. (Original post by ps1265A)
I still can't get my head around it... Based on what you've said, there twice as many moles of HCl than Ba(OH)2, but you need twice as much Ba(OH)2 to neutralize this 2 mols?
No, you need half as much barium hydroxide (in terms of moles) as HCl. You can see this from the stoichiometry of the reaction.
7. (Original post by charco)
No, you need half as much barium hydroxide (in terms of moles) as HCl. You can see this from the stoichiometry of the reaction.
Yes, I understand that. This should mean 2c1v1 = c2v2 HCl: Ba(OH)2, not c1v1 = 2c2v2

Each time I look at it it's changing. I know what you're trying to say, but I just need it in plain words. I'll give it another shot: does "2c2v2" mean that the number of moles of Ba(OH)2 is equivalent to HCl if it were to be timesed by 2?

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