VCE Chemistry Question Thread

[thbp]

The question says a 0.05M Na2CO3 solution for titration is prepared. Then it says: Calculate how much Na2CO3 is required to prepare 250ml of this solution.

Please help! What are we supposed to calculate (mole, grams?) what are the steps?

[chocolate.cake.1]

The question says a 0.05M Na2CO3 solution for titration is prepared. Then it says: Calculate how much Na2CO3 is required to prepare 250ml of this solution.

Please help! What are we supposed to calculate (mole, grams?) what are the steps?

Hey there,
All you have to do is plug those values into the formula n=cv
So you have the volume required, v=0.250 L (remember volume must be in litres for this equation)
And you have the concentration required, c=0.05 M

So n(Na2CO3) = 0.05 x 0.250 (sorry I don't have a calculator atm)
And you now can calculate the mass from this value using the n=m/M formula

[chansena]

Good Evening !!

Just a question regarding Volumetric Analysis (Determination of nitrogen content of law fertilizer)


Why does the unreacted sodium hydroxide = the same number of moles as NaOH?

Is that by pure coincidence? I'm a little bit confused, if someone could explain it that would be great

 Thanks !

[wunderkind52]

Good Evening !!

Just a question regarding Volumetric Analysis (Determination of nitrogen content of law fertilizer)


Why does the unreacted sodium hydroxide = the same number of moles as NaOH?

Is that by pure coincidence? I'm a little bit confused, if someone could explain it that would be great

 Thanks !

Hi!
Just a little confused with your question - is it a back titration?
And also, sodium hydroxide is NaOH

[chansena]

Hi!
Just a little confused with your question - is it a back titration?
And also, sodium hydroxide is NaOH

haha forgot to mention yeah it is a back titration and

haha wow i did not realize that typo its

Why does the unreacted sodium hydroxide  = the same number of moles as HCL ?

[keltingmeith]

haha forgot to mention yeah it is a back titration and

haha wow i did not realize that typo its

Why does the unreacted sodium hydroxide  = the same number of moles as HCL ?

Because you added enough HCl to match the amount of moles of NaOH.

Basically, you had excess HCl, but you stopped adding just after it reacted with all the NaOH.

[warya]

Didn't realise what page I was on and answered a question, disregard this post

[warya]

With dilutions, how do you know when to multiply the concentration by the the dilution factor or to divide? e.g. if I am taking a 10mL sample from a 100mL solution or vice versa?

Also how do you know when to divide by 10^3 etc to get from ppm to ppb 

[IndefatigableLover]

With dilutions, how do you know when to multiply the concentration by the the dilution factor or to divide? e.g. if I am taking a 10mL sample from a 100mL solution or vice versa?

Also how do you know when to divide by 10^3 etc to get from ppm to ppb 

Provided the number of moles stays the same, use the formula C₁V₁=C₂V₂. Substitute in the concentration which you currently know and the two volumes accordingly and then transpose to work out C₂ which would be your new concentration.
To know when you have to multiply the concentration by the dilution factor, draw a diagram of the actual setup so you can see clearly what is going on.

You'd actually multiply by 1000 (or 10^3) to get from PPM to PPB. You know that PPM is 1/10^6 and PPB is 10^9 (that is their differences) where they differ by 10^3. Simply multiplying your PPM by 1000 will yield it's PPB (E.g 4PPM x 1000 = 4000 PPB).

[warya]

Thank you!

Also, in IR spec, how come a molecule such as Hexane will produce a trough at C-H (2800-3000cm^-1) that is split, what results in the splitting of troughs in IR? I'm not confusing with NMR, however I am confused about why there isn't a single trough. And why are some troughs broad and others narrow?

In CH3COCH3 are the protons in the same environment? Text says they are however when I draw it, on CH3 group is bonded to the C in the double bond and the other to the O, doesn't this change the chemical environment for the H atoms?

[mahler004]

Thank you!

Also, in IR spec, how come a molecule such as Hexane will produce a trough at C-H (2800-3000cm^-1) that is split, what results in the splitting of troughs in IR? I'm not confusing with NMR, however I am confused about why there isn't a single trough. And why are some troughs broad and others narrow?

In CH3COCH3 are the protons in the same environment? Text says they are however when I draw it, on CH3 group is bonded to the C in the double bond and the other to the O, doesn't this change the chemical environment for the H atoms?

Yep, they are all in the same environment. You're not drawing it correctly (both methyl groups are bonded to the central carbonyl group - it's acetone.)

[lzxnl]

Thank you!

Also, in IR spec, how come a molecule such as Hexane will produce a trough at C-H (2800-3000cm^-1) that is split, what results in the splitting of troughs in IR? I'm not confusing with NMR, however I am confused about why there isn't a single trough. And why are some troughs broad and others narrow?

In CH3COCH3 are the protons in the same environment? Text says they are however when I draw it, on CH3 group is bonded to the C in the double bond and the other to the O, doesn't this change the chemical environment for the H atoms?

You've almost thought of it as dimethyl ether (CH3OCH3). The oxygen doesn't bond to the carbon here; it's more O=C(CH3)2
This means you have C=O, and the carbon makes its two other bonds by bonding to two methyl groups.

[Splash-Tackle-Flail]

In IR spec, it kinda makes sense that single bonds (e.g. C-C) would require less energy to move to a higher molecular energy level, than double/triple bonds (C=C) etc. But what I don't get is how come a higher molecular mass molecule doesn't need as much IR energy to 'stretch' than lower mass molecules? Wouldn't lower mass molecules be able to 'stretch' with less energy because there is less mass that needs to be excited?

[lzxnl]

In IR spec, it kinda makes sense that single bonds (e.g. C-C) would require less energy to move to a higher molecular energy level, than double/triple bonds (C=C) etc. But what I don't get is how come a higher molecular mass molecule doesn't need as much IR energy to 'stretch' than lower mass molecules? Wouldn't lower mass molecules be able to 'stretch' with less energy because there is less mass that needs to be excited?

In quantum mechanics, we model chemical bonds as springs. Then the restoring force is -kx and the potential energy is 1/2 kx^2, where x is the extension of the spring and k is the spring constant (just a constant). Classically, these springs vibrate at a particular frequency given by 1/2pi * sqrt(k/m) (I'm not going to prove that completely, but if you want to try, use F = -kx = ma, rearrange to get a = -kx/m, use a = d^2 x/dt^2 = v dv/dx to show that x is a sinusoidal function of t, notably x = A cos(sqrt(k/m)+constant))
The keen pedant will notice I used m instead of the reduced mass, which is actually what you should have in the above formula.

Using principles of quantum mechanics, it can be shown that quantum mechanical springs can only absorb light at the above frequency. So there are two factors at play here: the (reduced) mass of the system and the strength of the chemical bond (k is a measure of the strength of the spring). For heavier molecules, the mass increases and the frequency of oscillation (and hence of light absorbed) drops. This makes sense because for a given spring, heavier masses will accelerate slower.
For lower mass molecules, they now oscillate faster because of their lower mass.

You've misunderstood how IR works. It works not because of the energy required to stretch the bond; rather, it works because the frequency of light that can be absorbed is the same as the frequency of oscillation.

[Splash-Tackle-Flail]

In quantum mechanics, we model chemical bonds as springs. Then the restoring force is -kx and the potential energy is 1/2 kx^2, where x is the extension of the spring and k is the spring constant (just a constant). Classically, these springs vibrate at a particular frequency given by 1/2pi * sqrt(k/m) (I'm not going to prove that completely, but if you want to try, use F = -kx = ma, rearrange to get a = -kx/m, use a = d^2 x/dt^2 = v dv/dx to show that x is a sinusoidal function of t, notably x = A cos(sqrt(k/m)+constant))
The keen pedant will notice I used m instead of the reduced mass, which is actually what you should have in the above formula.

Using principles of quantum mechanics, it can be shown that quantum mechanical springs can only absorb light at the above frequency. So there are two factors at play here: the (reduced) mass of the system and the strength of the chemical bond (k is a measure of the strength of the spring). For heavier molecules, the mass increases and the frequency of oscillation (and hence of light absorbed) drops. This makes sense because for a given spring, heavier masses will accelerate slower.
For lower mass molecules, they now oscillate faster because of their lower mass.

You've misunderstood how IR works. It works not because of the energy required to stretch the bond; rather, it works because the frequency of light that can be absorbed is the same as the frequency of oscillation.

Ah, thanks!