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Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

1. For everyone who keeps getting 1.61 for the titration calculation

1) Calculate moles of S2O32- 0.2 x 20.15/1000 = 4.03x10^-3 moles
2) Use the mole ratios of the equation to figure moles of Cu2+ so Cu2+ moles=S2O32- moles
3) x10 to get the amounts of moles in 250 cm3 so =4.03x10^-2 moles
4) Calculate concentration by / (250/1000) =0.161 mol dm-3
2. I want this exam now!!
3. (Original post by CoventryCity)
For everyone who keeps getting 1.61 for the titration calculation

1) Calculate moles of S2O32- 0.2 x 20.15/1000 = 4.03x10^-3 moles
2) Use the mole ratios of the equation to figure moles of Cu2+ so Cu2+ moles=S2O32- moles
3) x10 to get the amounts of moles in 250 cm3 so =4.03x10^-2 moles
4) Calculate concentration by / (250/1000) =0.161 mol dm-3
the 3rd step isn't actually necessary cus the concentration will be the same in 25cm3 as it is in the 250 cm3, as long as you have the right amount of moles. jus sayin
4. (Original post by Flux_Pav)
oh thanks!
so you basically divide by how much there is in excess?
so say if u have a ratio thats 2:1 - acid:alkali would u multiply the excess amout by 2?
Okay say it was 0.5 H2S04 and 0.5 NaOH
you know that H2SO4 is in excess by 0.5, therefore you divide by the amount required to make one mole of water, so you have 0.5 moles of NaOH, and therefore 0.25 moles of H2SO4 is required
5. (Original post by haydyb123)
Okay say it was 0.5 H2S04 and 0.5 NaOH
you know that H2SO4 is in excess by 0.5, therefore you divide by the amount required to make one mole of water, so you have 0.5 moles of NaOH, and therefore 0.25 moles of H2SO4 is required
ohh ok so you have to use the acid always? not the alkali
6. (Original post by wilsea05)
the 3rd step isn't actually necessary cus the concentration will be the same in 25cm3 as it is in the 250 cm3, as long as you have the right amount of moles. jus sayin
I see, so you could just do moles in 25cm3 and divide by 25/1000
7. (Original post by volvicstar)
what is the equation for calculating H+ from a buffer system?
what is the general ionic equation for neutralisation?
what does a more negative electrode potential show?
outline how you would carry out a redox titration using MnO4-?
You need the concentration of the weak acid (undissociated) and the Ka value for that acid. Then you can work out the [H+] by rearranging the Ka equation. Then plug the [H+] into -log[H+] to find the pH

Neutralisation: Acid + Base (or alkali) --> salt and water
lets just use an alkali.
Full equation: HCl(aq) + KOH(aq) --> KCl(aq) + H2O(l)
Ionic Equation: H+(aq) + OH-(aq) --> H2O(l)
Things that dont change state or oxidation are removed from the equation to form an ionic equation.

More negative electrode potential means that it is more easily oxidised (loses its electrons) and a stronger reducing agent.

MnO4- mixed with a known volume of something...
eg. Fe2+ or something to do with that im not really sure
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) --> Mn2+(aq) + 4H2O + 5Fe3+ (aq)
Use Mol ratio of 5:1 Fe:MnO4-
8. (Original post by Flux_Pav)
ohh ok so you have to use the acid always? not the alkali
It depends what is in excess, I just used acid for simplicity
9. (Original post by Flux_Pav)
hey guys
came across neutralization and saw the page in the book
what if the number of moles of the acid and alkali are not the same? if that happens, which one of the two moles do you take?
and also, what if say 3 moles of water are being formed , what do you do then?
+ rep
I think you only need Q=MC(delta)T
and just use the mass of them both.
You use the equation to get mol ratios, but i dont think the mole ratios effect the result.
and what ever result you get from working out the MC(delta)T you have to then divide by 3 as you are making 3 mol of H2O, when Neutralisation is enthalpy change to make 1 mol H2O.
10. Hi guyss
Can anyone explaiin to me how redox titration equations are balanced?? for eg. how is H+ and balanced.

And also how do you write a two step equation reaction fast and slow step..

11. (Original post by Limesasquatch)
I think you only need Q=MC(delta)T
and just use the mass of them both.
You use the equation to get mol ratios, but i dont think the mole ratios effect the result.
and what ever result you get from working out the MC(delta)T you have to then divide by 3 as you are making 3 mol of H2O, when Neutralisation is enthalpy change to make 1 mol H2O.
I don't think you would ever get it as 3 moles of water forming as by definition, enthalpy of neutralisation is the formation of one mole of water.
12. (Original post by CoventryCity)
I see, so you could just do moles in 25cm3 and divide by 25/1000
yeah. if you have 250cm3 of a 1 mol dm3 solution, and take 25cm3 of that and put it in a beaker, that 25cm3 is still gonna be 1 mold dm3 unless you dilute it or something

the amount of moles is 10x less, and the volume is 10x less, which cancel each other out when you divide moles/volume so you get the same conc.
13. (Original post by CoventryCity)
For everyone who keeps getting 1.61 for the titration calculation

1) Calculate moles of S2O32- 0.2 x 20.15/1000 = 4.03x10^-3 moles
2) Use the mole ratios of the equation to figure moles of Cu2+ so Cu2+ moles=S2O32- moles
3) x10 to get the amounts of moles in 250 cm3 so =4.03x10^-2 moles
4) Calculate concentration by / (250/1000) =0.161 mol dm-3
OHHHHH!!! Thank you!! I calculated the concentration then times it by 10, when the concentration actually stays the samee.
14. (Original post by apo1324)
Umm I think CuCl4 2- is tetrahedral.

Ni(NH3)2Cl2 is square planar. Also cis-platin is square planar.

How do you distinguish if its a planer or tetrahedral cos they both have 4 coordinate bonds right??
15. (Original post by haydyb123)
It depends what is in excess, I just used acid for simplicity
in your example, wasn't the NaOH in excess ?

H2S04 + 2NAOH = NA2S04 2H20

so if u have 0.5 moles of acid and 0.5mol for the alkali, you know that the naoh is in excess so you divide 0.5 by 2 as it will form 1 mol of water

is it possible in neutralization for the no. of moles of acid and alkali to be different?
what would happen then?

LOL this part is doing my head in.
How do you distinguish if its a planer or tetrahedral cos they both have 4 coordinate bonds right??
Pt complexes are always square planar. i asked my teacher why, and she wrote a 5 page essay on it and handed it out to the class because i asked and she didn't know.

as a rule of thumb, if its got 4 coordinate bonds and isn't a Pt complex, its tetrahedral.
17. 25cm3 of a solution of CuSO 4.xH2O of concentration 26.50 g dm -3 was placed in a
conical flask and an excess of KI added. 22.25cm 3 of a 0.12 mol dm-3 solution of
sodium thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
Hi guyss
Can anyone explaiin to me how redox titration equations are balanced?? for eg. how is H+ and balanced.

And also how do you write a two step equation reaction fast and slow step..

Someone posted this in this thread, a few pages back,

How do you distinguish if its a planer or tetrahedral cos they both have 4 coordinate bonds right??
I think the rule is, everything other than Pt. complexes (with a co-ordination no. of 4) is tetrahedral, where as the Pt. complexes are square planar
20. haven't revised..does anyone have any revision notes on this module?

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