Stankovic123's chem q's

[zvezda]

Ahhh yeah thats what i was missing; the fact that the water used to dilute the solution has [H3O+] = 10^-7.
Cheers guys

[zvezda]

Hey,
Question on weak acids.
I did a q from VCAA 2008 exam 2 and it asked to "calculate the concentration of a methanoic acid solution that will have a ph of 2.1".
Going by the solutions, it technically asked for the equilibrum concentration, but how is this deduceable from first glance at the q? It didnt specify equilibrium conc. of the methanoic acid?
Help appreciated

[lzxnl]

The assumption is made for weak acids that the initial concentration IS the equilibrium concentration, which generally holds quite well for VCE questions.

From the Ka expression, Ka=[H+]^2/([HCOOH initially]-[H+]) if you want to be more accurate as the equilibrium concentration differs from the inital concentration by the amount of H+ formed
Sub in the H+ concentration from the pH value and you'll have your answer.

[zvezda]

The assumption is made for weak acids that the initial concentration IS the equilibrium concentration, which generally holds quite well for VCE questions.

From the Ka expression, Ka=[H+]^2/([HCOOH initially]-[H+]) if you want to be more accurate as the equilibrium concentration differs from the inital concentration by the amount of H+ formed
Sub in the H+ concentration from the pH value and you'll have your answer.

Ah ok.
Cheers nliu

[zvezda]

In UV/vis, what is actually detected by the detector?
Is it the energy that is emitted when electrons move back from higher energy levels after having been promoted or is it the energy that remains (ie the energy not absorbed from the light source)??
Thanks

[zvezda]

Why does a sample being analysed with a UV/VIS spectrometer need to be coloured?
Isnt there a detector that gives a readig of absorbances of UV light?

[zvezda]

These are from the 2009 VCAA exam 1:
1 - q7 in the multi choice. With the reaction between hydride and H2O, i understand that its a redox reaction by definition of change in oxidation number; however, technically there isnt a concrete donation/accepting of electrons. Isnt that what a redox reaction is?
So technically, the reaction should only be an acid-base reaction???

2 - q8 in the multi choice. The stem outlines that 14.80 mL of sodium hydroxide solution was required to reach the "endpoint" of a titration between a pure (0.132 gram) caboxylic acid and NaOH. When identifying the formula of the carboxylic acid, the solutions consider the "endpoint" as the equivalence point, which isnt correct??
With the calculations of the q, shouldnt we be finding the formula by deducing that less mol of NaOH actually reacted than what was added? Because an "endpoint" means past "equivalence".

Help is much appreciated

[lzxnl]

In UV/vis, what is actually detected by the detector?
Is it the energy that is emitted when electrons move back from higher energy levels after having been promoted or is it the energy that remains (ie the energy not absorbed from the light source)??
Thanks

It's the energy that is absorbed. Although really, you're quite right; the energy will be emitted later on as the electrons fall back down again.

Why does a sample being analysed with a UV/VIS spectrometer need to be coloured?
Isnt there a detector that gives a readig of absorbances of UV light?

It doesn't strictly need to be coloured I think for the reason you've given. It just makes identifying the appropriate wavelength for calibration easier.

These are from the 2009 VCAA exam 1:
1 - q7 in the multi choice. With the reaction between hydride and H2O, i understand that its a redox reaction by definition of change in oxidation number; however, technically there isnt a concrete donation/accepting of electrons. Isnt that what a redox reaction is?
So technically, the reaction should only be an acid-base reaction???

2 - q8 in the multi choice. The stem outlines that 14.80 mL of sodium hydroxide solution was required to reach the "endpoint" of a titration between a pure (0.132 gram) caboxylic acid and NaOH. When identifying the formula of the carboxylic acid, the solutions consider the "endpoint" as the equivalence point, which isnt correct??
With the calculations of the q, shouldnt we be finding the formula by deducing that less mol of NaOH actually reacted than what was added? Because an "endpoint" means past "equivalence".

Help is much appreciated

1. The hydride anion has two electrons, and when reacting with H2O, OH- forms and H2. What's happened?
Well, the hydrogen atoms in water, by the oxidation number rule, has zero electrons as the oxygen has grabbed them. After the hydride reacts with water, we're left with H2, where each hydrogen has one electrons. So, the hydride has donated an electron to the hydrogen in water. There IS a concrete donation and acceptance of electrons. If the oxidation number changes, redox MUST occur.
The converse is not always true. Redox can occur without a change in oxidation number. Write out a half equation for the reduction of ozone gas (O3) to form oxygen gas.
In acidic conditions, I'd get O3 + 2H+ + 2e- => O2 + H2O. Note how it is a reduction but oxygen in ozone and oxygen is zero. It's because there is an oxygen on the RHS with oxidation state -2 (also because of the funny bonding in ozone).

2. Endpoint is when the indicator changes colour. Generally an appropriate indicator is chosen so that the endpoint is close enough to the equivalence point. Remember the shape of titration curves, how steep they often get? If you used the wrong indicator, like methyl red for a reaction between ethanoic acid and sodium hydroxide, that would be bad.

[zvezda]

1 - so if the detector detects energy emitted by the electrons moving back down, why then in the VCAA 2009 exam 1 have they deemed this statement false: "the spectrum is a result of electrons falling back from higher to lower electronic energy levels"??

2 - wouldnt it have been a dangerous assumption to think that the endpoint would have been so close that the excess NaOH is negligible? They cant be interchanging different terms to give the same meaning.

Thanks nliu, always top quality responses

[lzxnl]

1 - so if the detector detects energy emitted by the electrons moving back down, why then in the VCAA 2009 exam 1 have they deemed this statement false: "the spectrum is a result of electrons falling back from higher to lower electronic energy levels"??

2 - wouldnt it have been a dangerous assumption to think that the endpoint would have been so close that the excess NaOH is negligible? They cant be interchanging different terms to give the same meaning.

Thanks nliu, always top quality responses

1. The detector technically doesn't; it detects the light that passes through the solution. What you see is absorbance, or how much light was absorbed by the solution. However, we all know that whatever light was absorbed went into promoting electrons into higher energy levels and that this light will be emitted. Does not mean that the detector picks up this light.

2. Is an indicator given? You can work out how good the assumption is from that.

[zvezda]

1. The detector technically doesn't; it detects the light that passes through the solution. What you see is absorbance, or how much light was absorbed by the solution. However, we all know that whatever light was absorbed went into promoting electrons into higher energy levels and that this light will be emitted. Does not mean that the detector picks up this light.

2. Is an indicator given? You can work out how good the assumption is from that.

Ah ok fair enough.
Nah no indicator is given... Lol

[lzxnl]

That's a dodgy question then

[Scooby]

What's the difference between AAS and UV-Vis in terms of which electrons absorb the light? I read somewhere that in AAS the valence electrons of the metal atoms absorb, whereas in UV-Vis it's the bonding electrons (then how do we analyse solutions of metal ions using UV-Vis?)

Also - is AAS restricted to metals or can we analyse other elements?

[lzxnl]

What's the difference between AAS and UV-Vis in terms of which electrons absorb the light? I read somewhere that in AAS the valence electrons of the metal atoms absorb, whereas in UV-Vis it's the bonding electrons (then how do we analyse solutions of metal ions using UV-Vis?)

Also - is AAS restricted to metals or can we analyse other elements?

I'm not sure of other differences, but here are some.
AAS is only used for atoms, where the electrons are in atomic orbitals. AAS promotes electrons to higher orbitals. This can be done conveniently using a vapour lamp of the same element as the element of interest. There was a VCAA question that used AAS for sulfur.
UV-Vis is used for molecular compounds. Here, the electrons are in molecular orbitals, orbitals formed by the overlap of atomic orbitals. These are slightly more complex than atomic orbitals and there are heaps more transitions possible.
Wiki says that UV-Vis can be used for metal ions, which makes sense. All UV-Vis consists of is excited electrons with light and seeing how much light of a given frequency was absorbed. I personally cannot see the difference in the actual chemistry that goes on with electrons being promoted to higher energy levels.

[zvezda]

Q19 on multi choice 2009 exam 2. Theres a solution of KI CaBr2 and both are at 1.0M. Now theres a electrolysis being carried out with inert electrodes. Apparently, iodide ions oxidise to iodine?? Doing this questions, i was thinking about how Cl- ions oxidise at high concentrations in preference to water, but this wouldnt occur with iodide would it because it is a much stronger reductant than water??
Help appreciated