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Titration curve/calculations help please! :(

a)

0.100 moldm^-3 NaOH(aq)
0.100 mooldm^-3 Ch3COOH(aq), pH = 2.9

50cm^3 of 0.1 moldm^-3 NaOH(aq) is gradually added to 25.0cm^3 of 0.1mol dm^-3 Ch3COOH(aq)

Sketch the titration curve and label the axes with approximate values of changes in pH

I've drawn a weak acid-strong base titration curve, the "vertical" bit going thru pH = 7

At what volume of NaOH would the titration neutralise? or is it just the amount added (25cm^3) as the concentrations of NaOH and CH3COOH are equal?


d)

Cpd B is an organic base.

A solution of B was prepared by dissolving 4.32g of B in water and making the solution up to 250cm^3.

A titration was carried out with 25cm^-3 of the solution of B, that was neutralised by exactly 23.20cm^3 of 0.200 HCl

1mol of B reacts with 1mol HCl. Calculate the Mr of B and suggest its identity

23.20cm^3(HCl) = 0.0232dm^3(HCl)

Moles(HCl) = 0.200 x 0.0232 = 0.00464.

Therefore 0.0464g = 1mol of B

Therefore 0.0464 = 4.32/Mr. Mr = 4.32/0.0464 = 93.103

This doesnt seem right :frown:

Could someone please help/correct me? :smile:

Thanks
Original post by Chemhistorian

Cpd B is an organic base.

A solution of B was prepared by dissolving 4.32g of B in water and making the solution up to 250cm^3.

A titration was carried out with 25cm^-3 of the solution of B, that was neutralised by exactly 23.20cm^3 of 0.200 HCl

1mol of B reacts with 1mol HCl. Calculate the Mr of B and suggest its identity

23.20cm^3(HCl) = 0.0232dm^3(HCl)

Moles(HCl) = 0.200 x 0.0232 = 0.00464. :smile:

Therefore 0.0464g = 1mol of B


whoa, where did this come from?

You have only taken 25ml from the original solution and found the moles of HCl that this reacts with.
If you have all 250 ml it would react with 0.00464 x 10 moles HCl = 0.0464

You are told that it's a 1:1 reaction, so the moles of base = 0.0464

AND you know the mass of this number of moles = 4.32g

So the RMM = 4.32/0.0464 = 93.1

So, same answer, although I don't see how you've got there... :confused:
Reply 2
Original post by charco
whoa, where did this come from?

You have only taken 25ml from the original solution and found the moles of HCl that this reacts with.
If you have all 250 ml it would react with 0.00464 x 10 moles HCl = 0.0464

You are told that it's a 1:1 reaction, so the moles of base = 0.0464

AND you know the mass of this number of moles = 4.32g

So the RMM = 4.32/0.0464 = 93.1

So, same answer, although I don't see how you've got there... :confused:


I have timesed by ten :smile:

Any idea on the base it could be? And what about the first question?

Thanks
Reply 3
Original post by charco
whoa, where did this come from?

You have only taken 25ml from the original solution and found the moles of HCl that this reacts with.
If you have all 250 ml it would react with 0.00464 x 10 moles HCl = 0.0464

You are told that it's a 1:1 reaction, so the moles of base = 0.0464

AND you know the mass of this number of moles = 4.32g

So the RMM = 4.32/0.0464 = 93.1

So, same answer, although I don't see how you've got there... :confused:


Also is it normal to find this bit of chemistry confusing? Its just Im good at maths but seem to struggle with the calculations bit :frown:

Thanks :smile:
Reply 4
I believe it could be aniline? C6H5NH2

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