woelen

hey i can't figure this question out...

 

"What is the pH of the 1.0x10^-8 M HCL?"

 

could i please have some help? Smile

 

Thanks

 

Sarah

For sake of brevity' date=' I introduce the symbol H for [H'] and OH for [OH].

For water we have H*OH = 10^-14

The acid HCl is a strong acid. At such low concentration, it can be assumed totally split.

Because for each OH(-) ion from water, we also have one H(+) ion, we now can say that H = OH + 10^-8, or OH = H - 10^8.

Now plug into the equation for water:

H * (H - 10^-8) = 10^-14.

H*H - 10^8 * H - 10^-14 = 0.

This is a quadratic equation in H. Solving this for H yields H = 1.05*10^-7.

Taking the -log10() of this yields 6.98, so the pH indeed equals 6.98.

The critical part of this computation is that at such very low concentrations, the contribution of the autoprotolysis of water cannot be neglected anymore.

Ahh yeah the O2 contamination makes sense now because it has to be produced when the water is split. How do you carefully control the conditions though? Put a lot of salt in and lower the voltage or what?

 

I don't understand why the H2 is formed at the cathode though (an Cl2 at the anode for that matter). I thought since the H2 was slightly positive due to its polar-covalence with 02 it would be attracted to the negatively charged anode. Can you explain this to me please?

 

edit: Doesn't the Oxygen combine in the liquid wqith the salt to form hydroxl ions?

If you have NaCl in solution, then this is split in ions Na(+) and Cl(-). The negative ions are attracted towards the anode, where they give off electrons and combine to Cl2. The positive Na(+) ions move towards the cathode. At the cathode, however, no metallic Na is formed, but hydrogen. The positive ions only take care of conduction of electricity, but it is the water, which is reduced.

If you want to produce Cl2 without too much O2 by means of electrolysis, then you need to use a graphite anode (copper dissolves) and a voltage, which is not too high.

a positive ion is called a cathode, while a positive electrodes is called the anide.

Positive ion: cation

Negative ion: anion

Positive electrode: anode

Negative electrode: cathode

"Just plain aqueous. No fusing at all..."

 

thats very' date=' very surprising. Fe(OH)3 is almost completely insoluble in water and i've NEVER seen KOH used as an oxidizing agent in any state except fused[/quote']

Well, I must say, I also never have seen KOH or NaOH as oxidizing agent . In fact, these are not oxidizing agents. What you are referring to is that KOH is used as supplement in order to facilitate a certain redox reaction (e.g. oxygen from the air, or KNO3 is the oxidizer and the NaOH or KOH supports the reaction).

 

Now coming back to your original question.

1) Fe(OH)3 is almost insoluble in water: This is true, but it does not prevent its oxidation. The metal zinc also is completely insoluble in water, yet it can be oxidized easily to Zn(2+) with just plain acid.

 

2) KOH (or NaOH) used as supplement to a redox reaction: This in fact is nothing special. At high schools, people are used to redox reactions, with H(+) as supporting reagent (e.g. Cr2O7(2-) or MnO4(-) as oxidizer in acidic environments), but the only reason for this is that the oxidizers, mentioned at high schools, usually contain a lot of oxo-groups, which must be neutralized with H(+). These common redox reactions use up H(+), and hence these reactions run best in acidic environments.

In the reactions, which I show in the experiment, it is the other way around. Look at the reaction equations for formation of ferrate or permanganate, and you'll see that these reactions use up OH(-), so these reactions run best in alkaline environments.

 

Some other examples of redox reactions, which require the ion OH(-) for their completion:

 

Oxidation of chromium (III), e.g. chromium hydroxide or a green solution of sodium chromite, with H2O2. In strongly alkaline aqueous solutions, you can easily oxidize chromium (III) to the yellow chromate with a little heating.

 

Oxidation of manganese (II) to manganese (IV) by oxygen from the air. This only happens in alkaline environments, because this reaction uses up OH(-).

is oxidation state in alot of cases just the ionic charge?

For simple ions: yes. Examples: oxide, chloride, bromide, sulfide, etc.

For more complex ions: no. For these situations you have to use some 'rules' for determining the oxidation state of all elements in such ions. Another poster has given some of these 'rules', together with exceptions. Oxidation state is a nice concept, but it must be used with care and you always have to be aware for pitfalls, when using this concept.

For budullewraagh: Metals can sometimes be in really weird oxidation states. The nitroso-complex of iron (produced in the brown ring test for nitrate, but also made by adding a nitrite to an acidic solution of a ferrous salt) has iron in its +1 oxidation state. The ion is [Fe(NO)](2+). The nitroso-ligand has formal charge +1, so the oxidation state of the iron must be +1.

What to think of rhenium, which can be brought in the -1 oxidation state, by adding zinc to an aqueous solution of a rhenium salt?

would any chemistry wizz's be able to help me out with this problem...?

 

i just don't see why the formal charge on the sulphur atom is zero' date=' i thought i would be +3.

 

Thanks

 

Sarah [/quote']

What do you mean with formal charge? Is this oxidation state? The oxidation state of the sulphur in the ion you show (sulfate) is +6. All oxygens have oxidation state -2. In reality, the charge distribution is not so dramatic. All oxygens have a slight negative charge and the central sulphur might have a certain positive charge. The total charge of the sulfate ion equals -2.

In your model, two oxygens are negatively charged, but in reality, all oxygens of the ion are similar and you cannot distinguish between them.

Its H2 at the anode but you can easily remedy this by separating the electrodes enough and placing a container directly above (or have the cathode penetrate the bottle) the cathode and collect the chlorine gas. ( the chlorine comes from the NaCl of the salt solution and the negatively charged Cl ion is attracted to the positively charged cathode where it combines with another Cl ion to form the stable Cl2). So, in effect you will get only pure Cl2 released above the cathode. Any other contaminates will be non-gaseous and therefore wont dillute your Cl collection. I think the hyrdrogen is released from the anode because of its strong electronegativity... This is my own personal understanding based on my own independent reading so please correct me if I am wrong on any of this.

I think you exchanged the role of anode and cathode. Cl2 is formed at the anode and H2 is formed at the cathode.

Another point is that you WILL get gaseous contaminants. At the anode you also will have formation of oxygen. Formation of chlorine gas and formation of oxygen are two competing reactions. The conditions must be carefully controlled in order to minimize the production of oxygen at the anode.

Ok I got some car battery sulfuric a while ago and today I needed some hydrogen so I dumped a nail it 25ml. Well after I got the Hydrogen I needed I decided to leave the nail in there to see how much it would disolve. Here I am about 6 hours later and the nail is still moderately reacting. How concentrated do you think this acid is? The nail was not galvanised and it has been sitting at room temperture all day. I did some very rough calculations when I first got it and worked it to be about .4mol/L althought this calc was done rushidly and poorly. I can't be arsed titrating it so can anyone give me an idea on how conentrated you think it will be?

 

~Scott

Do not let the speed of the reaction mislead you. Many metals, which react with acids, according to the electropositive series, do so very slowly. I have tried with iron, chromium, titanium, tin, etc. All react very slowly, even in 30% HCl or 30% H2SO4. Even zinc is not that fast. The only real fast one I found is magnesium and with some patience, aluminium also can be made to react fast in concentrated acids. Of course the alkali metals also are really fast.

Oxidation state is just a tool for bookkeeping of electrons. Bookkeeping is done, such that it gives a good impression of how far an element is oxidized (or reduced).

Some elements gain electrons easily (they act as oxidizers and they are reduced). The most common is oxygen. The oxo-group (or oxide ion) is said to have oxidation state -2. Hydrogen usually has oxidation state +1 in its compounds (except in metal hydrides, where it has oxidation state -1). Using these simple rules, one can easily compute the oxidation state of other elements.

Example: MnO4(-). Total charge: -1. Charge per oxo-group: -2. So, you need +7 for manganese in order to get a total charge, equal to -1.

Keep in mind though, that in MnO4(-), the manganese does not really have a charge equal to +7. In many real life compounds, the charge is not distributed as extremely as oxidation numbers would suggest. So, in reality, the charge on the manganese will be just a little over 1 and the charge on each of the oxo-groups is a little below 0, such that the total charge still remains -1.

The usefulness of oxidation numbers is in that they easily tell you whether an element is strongly oxidized or not. In my experiment, you have Fe(OH)3, with Fe in the +3 oxidation state. In the ferrate ion, the iron has formal oxidation state +6. So, from this playing with numbers, you immediately see, that the ferrate is an iron-species, which is oxidized further than ferric hydroxide.

Sometimes the concept of oxidation number is flawed. Consider the deep blue compound CrO5. This does not contain chromium in the +10 oxidation state, but it contains chromium in the +6 oxidation state, with an oxo-group attached to it and two peroxo-groups attached to it. For one oxygen atom we have oxidation state -2 and for the other four oxygen atoms we have oxidation state -1. A better formula is CrO(O2)2. So, with oxidation numbers you have to be careful and you have to know your compounds.

Just another nice one: try to determine the oxidation state of S, C and N in the thiocyanate ion SCN(-) and in the thiosulfate ion S2O3(2-). Here you'll also see that the concept of oxidation number sometimes is not that easy.

woelen' date=' on your site the reaction:

2Fe(OH)3 + 3ClO– + 4OH– → 2FeO42- + 3Cl– + 5H2O

is a bit unclear to me. surely the hydroxide must be fused, not aqueous, right?[/quote']

Just plain aqueous. No fusing at all...

also how in world is this meso?! aghhh stress...!!

I assume that this picture is 1,2-cyclohexane-diol (1,2-dihydroxy cyclohexane). In this molecule, the bonds between the C-atoms are single. This means that the two items, connected at the C-atoms are pointing downwards and pointing upwards. So, you can have that both OH-groups are pointing upwards (relative to the plane, in which the 6-membered-C-ring is) or both of them are pointing downwards, or they point in opposite directions. Because the C-atoms are in a ring, there is no free rotational motion, so, a OH-group pointing downwards will remain so forever (assuming that the molecule as a whole is kept in the place and not rotated around an axis in the plane). In this way, just from plain 3D-geometry, you can easily see that a molecule is meso or not. In the picture you give, both OH-groups are pointing upwards.

are you sure? because my answers booklet says that they are the same

Indeed, they are the same, they are not enantiomers. The drawing of molecules like these in 2D is somewhat confusing. Imagine the right picture, and rotate the molecule, such that the H is pointing upwards. Then you have CH3 coming out of the plane, towards you, Br in the plane and NH2 going out of the plane, pointed away from you. This is the same situation as in the left picture.

Today, I did another nice experiment, now with selenium. I transformed the black allotrope into the red allotrope. Less risky than the chromyl chloride experiment, but fun also . This may also be interesting for people who collect elements.

Black selenium is affordable. It can be obtained for just over $10 per 30 grams, including shipping worldwide from http://www.emovendo.net .

This is sufficient for all the experiments with selenium you can ever dream of . For my experiment you only need approximately 10 mg, so you have some left for other experiments as well :D.

For the experiment, see my site again:

http://81.207.88.128/science/chem/exps/selenium/index.html

Have fun and stay green.

ivew found some oxalic acid, which is prertty strong for an organic acid. i was wondering if i could make some other oxalate or somekind of other thing that wouldnt be an oxalate anymore, or a ester, or someting- something, that i could test to make sure i got or if i coukd use it for sometin. any thoughts?

Oxalate forms beautiful complexes with iron and chromium. Mix a solution of oxalic acid with a solution of ferric chloride or ferric sulfate and then add some sodium hydroxide.

You'll get a nice green complex, trisoxalato ferrate (III), [Fe(C2O4)3](3-). The color of the complex is shown here in the form of its ammonium salt:

http://81.207.88.128/science/chem/compounds/ferric_amm_oxalate.html

With your acid and a ferric salt, you can make this complex yourself. You can even isolate the potassium salt of this, because that is only sparingly soluble.

If you do the same with a ferrous salt, then you get a deep yellow complex, bisoxalato ferrate (II), [Fe(C2O4)2](2-).

A similar experiment you can do with chromium. Add some oxalic acid to a solution of potassium dichromate and also add a small amount of dilute hydrochloric acid or sulphuric acid. Then slightly heat. The liquid will turn deep purple. This is an oxalato complex of Cr(3+). I do not know its precise formula though.

wow u are smart, how do u come up with experiments like this? (great way to continue my inorganic exp.'s)

Buy a good old book on chemistry. Pre-war books are best. These books describe all kinds of compounds, which can be made with fairly common chemicals. Especially books in the German language are very nice. Apparently before WW II, the German chemists were focussing more on compounds and processes.

I have books with 1000+ pages from 1920 and 1915 and these describe all kinds of compounds. With these books at hand and a little reasoning you can come up with many cool experiments.

Modern books focus much more on theory, which on its own is OK, but sometimes they tend to go too far with this. Just knowing the basic properties of common compounds makes chemistry much more fun.

really toxic and carcinogenic.

 

good idea though; sounds like an interesting compound. i'm half tempted to make some' date=' but the gas is so dense and such a strong oxidizing agent that any inhalation would probably prove fatal[/quote']

Well, it is toxic (carcinogenic), but a small whiff of this gas won't kill you. Of course you should do your best to not inhale any of this gas, but if you do inhale something, then it is not fatal. In fact, I accidently could smell some of the gas after taking the picture with the gas in front of the bottle in the air and I left the place immediately, but I'm still here to show you the results. Remember, I only used approximately 100 mg of K2Cr2O7, so the whiff I inhaled probably will be in the microgram range. The smell actually is quite strange: sweet and spicy.

Mmm, very nice experiment. I'm certainly going to try it ! I'll let you know if it worked well or not. But I was wondering. Aren't there nice experiments with chromyl chloride for example ?? Would be nice

Yes, you can do nice things with chromyl chloride. You can dissolve this stuff in relatively inert organic solvents, such as ligroin (Dutch: 'laagkokende wasbenzine' / 'petroleumether'). The solution in ligroin then can be used for other experiments, such as oxidation of organic compounds in non-aqueous solvents. The stuff, however, should not be stored, as it is quite unstable and with some organics, it forms explosive mixtures or may cause fire. It also eats almost everything. It is even worse than bromine on storage. I kept it in the little bottle for just a few hours and in that short time, the plastic screw-cap already was severely corroded.

You can make a solution in ligroin, simply by adding some ligroin to the mixture. With a glass pipette, you can carefully take away the ligroin layer, with some of the CrO2Cl2 dissolved in it. The ligroin remains above the sulphuric acid layer.

I want to share with you a nice experiment with a colored gas, which contains chromium. It looks like bromine, but in reality it is a very volatile metal-compound!

http://81.207.88.128/science/chem/exps/volatile_chromium/index.html

Have fun. If you repeat the experiment, please be very careful. The gas is really toxic!

That is interesting about ending up where you start from' date=' I see what you mean. But I am trying to visualise the sphere still.

 

If a hyper-plane was cutting the 3-sphere slowly from one end to the other. Me being in the hyper-plane would see a regular sphere appear, grow and then reach a maximum and shrink back down into a point?[/quote']

Yes, you would see a regular sphere (hollow ball) appearing out of 'nothing', growing, shrinking back and disappearing again.

Does it make sense to say that two points is a zero-sphere?

Yes, it does. A countable set of points is said to have dimension 0.

I read that an n-spehere is the structure made from all the points being exactly the same radius from a single point in (n+1) dimensions (thats of the top of my head). But also shouldnt the resulting structure be of n-dimensions itself? I'm not sure whether two points would be called 0 dimensional would it?

An n-sphere indeed is a (curved) n-dimensional space of finite n-volume. For 1-dimensional spaces (a circle), the 1-volume is the length of the circle, for 2-dimensional spaces (an ordinary sphere, the surface of a ball), the 2-volume is the surface of the sphere, etc. The n-volume of an n-sphere is proportional to r^n, with r being the radius of the n-sphere, e.g. the 1-volume of a circle is 2*pi*r, the surface of a sphere equals 4*pi*r^2. For a 0-sphere, the 0-volume does not depend on the radius, it can be written as k*r^0, being a constant k.

Also, could one visualise a 3-sphere (picturing it with 3 spatial dimensions and the fourth as time) as a point that grows to a sphere and then shrinks back down?

A three-sphere has nothing to do with time. It simply is a geometrical object. A nice way to visualize a 3-sphere is the following: Suppose you are on the sphere floating around in space (this is imaginable, because you are 3D as well I hope ) and start moving forward. If you keep on moving forward in a "straight line" in the same direction, then you'll eventually reach your original position again.

If you cannot imagine this, then think of the analogon of a 2-sphere with you being on the surface (of e.g. a planet). If you walk to a certain direction and you keep on moving, then you'll end up at your initial position again.

I place the words "straight line" between quotes, because of the fact, that such thing not really exists on a sphere. A sphere is not an euclidian space, although locally it approaches an n-dimensional euclidian space. In curved spaces, the concept of straight line must be replaced by the more general concept of "geodesic". Google is your friend on this subject.

does anyone know some interesting synthesis/experiments with organic compounds? and im just a beginner so i dont have much knowledge or experience with this field. thx

What organic compounds are you referring at. Beware, organic chemistry is not the easiest to deal with. Organic reactions frequently are slow, incomplete and frequently not spectacular to see (input is colorless and output still is colorless). If you want to appreciate organic chemistry, then you really have to know what you are doing, otherwise you'll be disappointed quickly!

Besides what I mentioned above, organics frequenly involve volatile solvents and reactants. This introduces a severe additional risk for your health.

If I were you, stick to inorganics for the time being. This is easier, the chemicals can be obtained more easily and the risks for your health are less (there are some exceptions, but the average home chemist will not come across these usually).

ahaha i see. but in compounds i guess u could consider those halogens as having radicals, (kinda, not really tho)

You cannot say that a certain atom (e.g. the halogen) has a radical, a compound or ion as a whole is a radical (or not).

For the halogens, the compound ClO2 is regarded a radical, but Cl2O is not. Try to draw a Lewis structure of ClO2 and you'll see that you have trouble making such a model, while it can be done easily for Cl2O. The same difficulty exists for NO2 and NO, but not for N2O5, N2O3 and NO3(-).

Hello,

Suppose we have a truncated taylor series for exp(x), truncated at the

term, just beyond the N-th power. Let's call these truncated series

trunc(N, x). Some examples:

trunc(0, x) = 1

trunc(1, x) = 1 + x

trunc(2, x) = 1 + x + x^2/2

...

trunc(N, x) = 1 + x + x^2/2! + .... x^N/N!

Here N! means the factorial of the integer N.

Here, trunc(N, x) is an N-th degree polynomial over the complex numbers

with variable x.

The zeros of trunc(N, x) are on a very regularly shaped curve. The

shape of the curve hardly depends on N.

As an example I'll show the zeros for trunc(400, x) as a gif image,

computed by means of my polynomial solving routine (which uses adaptive

multiple precision arithmetic).

http://www.woelen.nl/zeros-trunc400-exp.gif

The image shows the distribution of the zeros of trunc(400, x) in the

complex plane. Each red dot represents a single zero. As the image shows, the zeros are nicely distributed along a curve.

My question is, is there a closed analytic expression for the curve, as

function of N, or is there a closed analytic expression for the

limiting curve, for N going towards infinity?

When a plot is made for other values of N, then the curve looks very

similar. For increasing N, the curve tends to blow up, but its basic

shape hardly changes.

If some of you has any idea about an expression for the curve, then I

would be pleased to read about that.

Thanks,

Wilco

PS: Source code of the polynomial equation solver, together with some test programs is available at http://woelen.scheikunde.net/science/software/mpsolve.html

so valency is the # of electrons on the valence shell

What is the valence shell? This concept works quite OK for e.g. sodium, chlorine, etc. For transition metals it does not work anymore. Many transition metals use electrons from their outer shell, but also from deeper shells in their chemical reactions.

For the lantanides things get even more complex. With these, the second deepest shell also is involved in the chemistry of the elements, so one cannot simply speak of THE valence shell. As I stated already, for these elements, the only way to understand how it is is by performing mathematical simulations of quantum mechanical models.

This is very interesting' date=' thankyou for your knowledge woelen. These insights have sparked several other questions I am pondering. Is it possible for something to have a valency of between two integers [eg 0.5+'] or do we just say it is a particularly unreactive particle of valency 1 ?

 

More Q but midnight here, cya tomorrow!

From this, I understand that you relate valency to reactiveness. You mean that the higher the valency, the higher the reactiveness?

The fact is that valency is not related to reactiveness. The element fluorine has valency 1 and has no other possibilities. Yet, it is the most reactive element there is and also among the compounds (which is a much broader class) it is among the most reactive ones. On the other hand, an element as sulphur frequently has valency 2, but it is not really reactive.

The concept of fractional valency I have never seen. Remember, the concept of valency is introduced to make reasoning about chemical compounds and bonds somewhat easier, but with many real-life molecules this concept gets flawed and more advanced concepts like molecular orbitals or even low-level quantum mechanics is needed in order to understand the structure of the compound.