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# Chemistry Titration calculation watch

1. will upload in a min

2. part b

3. The initial mixture contains two bases, CO32- and HCO3-.

The CO32- reacts first (forming extra HCO3-). The first end point is reached when all of the CO32- is reacted.

The second end point is reached when [the HCO3- that was present at the start AND the extra HCO3- that was created in step 1] has reacted.

Do the math.
4. (Original post by Pigster)
The initial mixture contains two bases, CO32- and HCO3-.

The CO32- reacts first (forming extra HCO3-). The first end point is reached when all of the CO32- is reacted.

The second end point is reached when [the HCO3- that was present at the start AND the extra HCO3- that was created in step 1] has reacted.

Do the math.
here is my working
so 12.45 moles HCL react with HCO3

C V(HCO3) =C V(HCL)
C X V = 0.150 X 12.45/1000

what will be the volume of HCO3
initial volume was 25 but in step 1 we also added 20.80 HCL so Volume will be 25+20.80?
5. (Original post by Qer)
here is my working
so 12.45 moles HCL react with HCO3

C V(HCO3) =C V(HCL)
C X V = 0.150 X 12.45/1000

what will be the volume of HCO3
initial volume was 25 but in step 1 we also added 20.80 HCL so Volume will be 25+20.80?
12.45 would be a volume, i.e. cm3 not moles.

I would use the 20.80 to work out how many mol of CO32- there are (and therefore how many mol of HCO3- made)
I would use the 33.25 to work out the total number of mol of HCO3- there are.
I would do a subtraction to work out how many mol of HCO3- there were initially.
BUT, doing the subtraction first is equally valid.

The 25 is the volume of buffer and has nothing to do with the 20.80, which is a volume of HCl. The conc of the HCO3- initially is n(HCO3-)/(25/1000).
6. (Original post by Pigster)
12.45 would be a volume, i.e. cm3 not moles.

I would use the 20.80 to work out how many mol of CO32- there are (and therefore how many mol of HCO3- made)
I would use the 33.25 to work out the total number of mol of HCO3- there are.
I would do a subtraction to work out how many mol of HCO3- there were initially.
BUT, doing the subtraction first is equally valid.

The 25 is the volume of buffer and has nothing to do with the 20.80, which is a volume of HCl. The conc of the HCO3- initially is n(HCO3-)/(25/1000).
that makes sense
But I was confused
Yoy mentioned that it is buffer and has nothing to do with the 20.80, which is a volume of HCl.

but HCL is reacting to buffer then why it has no effect on that
7. There are three numbers in the Q:

25 cm3 - the volume of the CO32-/HCO3- buffer.
20.80 cm3 - the volume of HCl required to neutralise just the CO32- in the buffer
54.05 cm3 - the total volume of HCl added, that neutralises the CO32- first (creating some HCO3-) and then all of the HCO3-.

Once you work out how many mol of HCO3- there was initially, you're asked to work out the conc. To do that, you need the 25 cm3 - the volume of the buffer.
8. (Original post by Pigster)
There are three numbers in the Q:

25 cm3 - the volume of the CO32-/HCO3- buffer.
20.80 cm3 - the volume of HCl required to neutralise just the CO32- in the buffer
54.05 cm3 - the total volume of HCl added, that neutralises the CO32- first (creating some HCO3-) and then all of the HCO3-.

Once you work out how many mol of HCO3- there was initially, you're asked to work out the conc. To do that, you need the 25 cm3 - the volume of the buffer.
THANKS
9. equ for the reaction between CH3COCL and CH3CH2NH2

where CHCH2+ come from?

Isn't it should be CH3CH2NH3CL?
10. CH3CH2NH3Cl (there is a 3 missing) is called ethylammonium chloride. The + is on the N in NH3, not on a C.

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