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During my class experiment i carried out this experiment:

Method:
Page 1 of Methodhttp://i46.tinypic.com/2iau43l.jpg
The absorption values i found out to be:
1      0.104
2      0.195
3      0.234

and so on.

Page 2 of Methodhttp://i47.tinypic.com/mjque8.jpg

My problem: Calculating the concentration of PNPG, and then going on to calculate the initial reaction velocity.

My lecturer said this which i dont understand:  We are told to work out the new concentration of PNPG in the 10 ml mixture. We are given the concentration of PNPG, the amount of liquid removed and the new volume it is introduced to, hence you need to work out the new concentration of PNPG in the mixture.

From a previous calculation we discovered a conversion factor 333.33 nmole/ml ( our lecturer said to use this value for something) and this was used to calculate the concentration of PNP by multiplying the converting factor by the absorption.

I would appreciate any help given.

Thank you.

[Dan]

My lecturer said this which i dont understand:  We are told to work out the new concentration of PNPG in the 10 ml mixture. We are given the concentration of PNPG, the amount of liquid removed and the new volume it is introduced to, hence you need to work out the new concentration of PNPG in the mixture.

Ok, you have a stock PNPG solution that has a concentration of 0.003 M (M = mol/L), and you're essentially diluting it. Since you know how much of the stock solution you are diluting, and the final volume of the new solution, you can calculate the new concentration.

Consider that: concentration (M) x volume (L) = number of moles (mol)

so if you have your initial concentration c_i, final conc c_f, and volumes V_i and V_f, since the number of moles of PNPG is constant:

c_iV_i = c_fV_f

Rearrange and crunch the numbers...

[gggggggggg]

My lecturer said this which i dont understand:  We are told to work out the new concentration of PNPG in the 10 ml mixture. We are given the concentration of PNPG, the amount of liquid removed and the new volume it is introduced to, hence you need to work out the new concentration of PNPG in the mixture.

Ok, you have a stock PNPG solution that has a concentration of 0.003 M (M = mol/L), and you're essentially diluting it. Since you know how much of the stock solution you are diluting, and the final volume of the new solution, you can calculate the new concentration.

Consider that: concentration (M) x volume (L) = number of moles (mol)

so if you have your initial concentration c_i, final conc c_f, and volumes V_i and V_f, since the number of moles of PNPG is constant:

c_iV_i = c_fV_f

Rearrange and crunch the numbers...

So , the initial conc. is 0.003M X 6 ( is this the dilution value)? = 0.018 M
Final Volume = 0.01 L
Initial volume = 0.0001 L
Final Conc = 1.8 M  ?

[Borek]

So , the initial conc. is 0.003M X 6 ( is this the dilution value)? = 0.018 M

So, after DILUTION your concentration is higher than initial? And it doesn't sound an alarm bell for you?

[gggggggggg]

So , the initial conc. is 0.003M X 6 ( is this the dilution value)? = 0.018 M

So, after DILUTION your concentration is higher than initial? And it doesn't sound an alarm bell for you?

O dear. I get what you mean.

Is it 0.003 M / 6 =  5^-4
then 5^-4 X 0.01 L = 5^-6
then 5^-6/0.0001 L = 0.05 ?

Or is it 0.003 X 333.33nmole/ml ( Used the bear lambert law to calculate a converting factor which includes the dilution)
then (0.999 X 0.01 L) 0.0001 L = 99.9 nmole/ml

[Dan]

I don't understand what you're doing in your calculations.

c_iV_i = c_fV_f

Rearrange:

c_f = c_i(V_i/V_f)

c_f is the final (or new) concentration - which is unknown. The terms on the right hand side of this equation are all given, crunch the numbers...

As you are diluting, you should expect c_f<0.003 M

Example:

1 mL of 1 M HCl is diluted to a final volume of 20 mL, find the new concentration.

c_i = 1 mol/L
V_i = 0.001 L
V_f = 0.020 L

c_f = c_i(V_i/V_f)
c_f = [1 mol/L] x ([0.001 L]/[0.020 L])
c_f = [1 mol/L] x 0.05 (I think this is the dilution factor you refer to)
c_f = 0.05 mol/L (or 0.05 M)

[gggggggggg]

I think i understand.

Is this right?

Ci= 0.003M
Cf=?
Vi=0.0001L
Vf=0.010 L

Therefore ....Cf= 0.00003 M

Then 0.00003 X 6 =1.8^-4

So, 1.8^-4 X 10^6 = 180 nmole/ml                     

Is this correct or am i still a total dummy.

[Dan]

Therefore ....Cf= 0.00003 M

I agree

Then 0.00003 X 6 =1.8^-4

Perhaps I am missing something... Can you explain why you are multiplying your final concentration by 6?

[gggggggggg]

Therefore ....Cf= 0.00003 M

I agree

Then 0.00003 X 6 =1.8^-4

Perhaps I am missing something... Can you explain why you are multiplying your final concentration by 6?

"For the conc of substrate I know that  0.003M is for 1000ml, i have to work out moles for 0.1 ml and this is the number in 10 ml . then  scaling up to 1L." - what my lecturer said.

Can you explain this please.I get the part about 0.003 M is for 1000ml. Then i have to work out moles for 0.1 ml, which is 0.1X0.003M = 0.0003 moles/ml    then i just do 0.0003 moles/ml X 1000 to get = 0.3 moles/L

Is this what my lecturer means?

[Dan]

This the same idea I have shown you. You have already calculated the correct answer, but then multiplied it by 6 for reasons that are not apparent to me. Let's look at this...

"For the conc of substrate I know that  0.003M is for 1000ml, i have to work out moles for 0.1 ml

Ok, so you have to work out the number of moles (N) in 0.1 mL (= 1 x 10^-4 L), given there are 0.003 mol in 1 L.

concentration = amount/volume

c_i = N/V_i

so,

N = c_iV_i

and this is the number in 10 ml .

Yes, you have N moles in 10 mL (= 1 x 10^-2 L).

concentration = amount/volume

c_f = N/V_f

c_f is the answer you are looking for.

If you combine the equations above, you get c_f = c_i(V_i/V_f) as I have posted before

then  scaling up to 1L."[/b] - what my lecturer said.

Ah, OK, I would have worked with all volumes in L from the start so this wouldn't be necessary. If you work with mL all the way through (this requires converting concentrations to mol/mL as well), then you need to convert to L at the end.

Then i have to work out moles for 0.1 ml, which is 0.1X0.003M = 0.0003 moles/ml

Keep all your volumes in litres - you are mixing units of volume, which is giving you the wrong answer.

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Thank you so much for the help m8.