Question Chain Thread !!! (1 Viewer)

[danz90]

doesn't the neutralisation depend on the [H₃O⁺] ?

For any acid, weak or strong, if they are both pH 3 , then they must have equal [H₃O⁺] , yes their acid concentrations will be unequal... but doesn't neutralisation depend on the [H₃O⁺] ??

Confucious lol

 

[Trebla]

doesn't the neutralisation depend on the [H₃O⁺] ?

For any acid, weak or strong, if they are both pH 3 , then they must have equal [H₃O⁺] , yes their acid concentrations will be unequal... but doesn't neutralisation depend on the [H₃O⁺] ??

Confucious lol

I agree. They should be equal amounts of alkaline, to neutralise equal amounts of H₃O⁺. However, in terms of calculations, you cannot determine the concentration of acetic acid unless you know its degree of ionisation.

Anyways, I found this rather challenging question you might want to take a look at lol:


A student is trying to work out the formula of an unknown neutral acid he was given. So far he knows that the formula takes the form H_nXO₃, where n is an unknown positive integer and X is an unknown non-metallic element which can either be Phosphorous (P), Sulfur (S), Carbon (C) or Nitrogen (N). To determine which of the four elements X actually is for the unknown acid, the student performs a series of titrations.

The student decides to prepare a secondary standard solution of NaOH by titrating it with a primary standard solution of 0.1064 mol/L of HCl.

The result was that 25.00mL of NaOH required 25.00mL of HCl with concentration 0.1064 mol/L for an end-point with the indicator colour change.

The student then dissolved 2.569 grams of H_nXO₃ in water to make a 250.0mL solution. He performed a titration with the unknown acid H_nXO₃ and the standard NaOH solution.

The result was that 25.00mL of the standard NaOH required 16.31mL of H_nXO₃ solution for an end-point with the indicator colour change.

Determine the identity of X and the value of n. Justify your answer.

 

[Undermyskin]

danz, wake up!

When you have 0.1 M acetic acid, and 0.001 M HCl, which one needs larger volume of NaOH 0.1 M to neutralize? (taking same volume for both acids) Acetic acid.

Using a simple calculation when we have 1L of each acid.

n CH3COOH = 0.1 * 1 = 0.1

n HCL = 0.001 *1 = 0.001

Neutralization is summarized into
H30+ + OH - --> 2H2O

which can be confusing when calculating. But, um, let's put it this way: ionization of CH3COOH is an equilibrium system. Consuming the H+ pushing the reaction to the right in order to replace the lost H+. By continuing in this direction, eventually all potential H+ is consumed. As you can see, more NaOH is needed.

Ta ra!

Trebla, you're ignoring the fact weak acids do not ionize completely and they form equilibriums. Plus, to neutralize an acid by a base means completely reacts with all H+ present and potentially present.

PS: your question is evil.

nNaOH + HnXO3 --> NanXO3 + nH2O

nNaOH = 0.1064 * 0.025 = 2.66 * 10^-3

--> # mole of HnXO3 = 2.66 * 10^-3/n

[HnXO3] = 2.66 * 10^-3/n* 16.31 * 10^-3 = 38/233n ......(1)

However, [HnXO3] = 2.569/[1.008n + Mx + 48]0.25 ........ (2)

(1) = (2)
--> a few maths steps
--> 62n = Mx + 48

By trying, n = 1, X is N. (yay!) Mx is the atomic mass of X.

Q: Assess the impacts on the environment of the developments in technology available to manufacture NaOH. (from CSSA either 05 or 06)

 

[danz90]

danz, wake up!

When you have 0.1 M acetic acid, and 0.001 M HCl, which one needs larger volume of NaOH 0.1 M to neutralize? (taking same volume for both acids) Acetic acid.

Using a simple calculation when we have 1L of each acid.

n CH3COOH = 0.1 * 1 = 0.1

n HCL = 0.001 *1 = 0.001

Neutralization is summarized into
H30+ + OH - --> 2H2O

which can be confusing when calculating. But, um, let's put it this way: ionization of CH3COOH is an equilibrium system. Consuming the H+ pushing the reaction to the right in order to replace the lost H+. By continuing in this direction, eventually all potential H+ is consumed. As you can see, more NaOH is needed.

Ta ra!

But that's different from saying that they BOTH have a pH of 3. You can't actually determine the [H₃O⁺] of a weak acid, because you don't know to what extent it has ionised. Whereas, with a strong acid, you can, because virtually all acid molecules ionise.

If the solutions both have a pH of 3 then they must have equal [H₃O⁺] , regardless whether they are weak or strong, going by the rule: pH = -log₁₀ ( [H₃O⁺] )

Of course, if there are two solutions of equal concentration, lets say 0.1M, one acetic and the other HCl, then of course the HCl would require more alkaline solution than the acetic acid, because the HCl will ionise more extensively producing more H₃O⁺ that need to be neutralised.

 

[Undermyskin]

What?

OK, define 'neutralization'.

 

[minijumbuk]

But that's different from saying that they BOTH have a pH of 3. You can't actually determine the [H₃O⁺] of a weak acid, because you don't know to what extent it has ionised. Whereas, with a strong acid, you can, because virtually all acid molecules ionise.

If the solutions both have a pH of 3 then they must have equal [H₃O⁺] , regardless whether they are weak or strong, going by the rule: pH = -log₁₀ ( [H₃O⁺] )

Of course, if there are two solutions of equal concentration, lets say 0.1M, one acetic and the other HCl, then of course the HCl would require more alkaline solution than the acetic acid, because the HCl will ionise more extensively producing more H₃O⁺ that need to be neutralised.

True. At the time they were initially measured, both strong/weak acid had the same amount of H3O +.
But the amount of H3O+ in HCl is already the maximum, right from the start, since it's complete ionisation.
However, for CH3COOH, there were molecules of CH3COOH that remain undissociated. So during the titration, the H+ ions get used up, and the CH3COOH shifts to dissociate to form more H+ ions, hence a bigger volume of NaOH required for titration.

 

[adnan91]

What?

OK, define 'neutralization'.

A reaction between a base and an acid taken out usually by titration.

Equivalence point (neutralization point) : The point at which the concentration of H+ ions = OH- ions

End point: The point at which the indicator changes colour

 

[Undermyskin]

True. At the time they were initially measured, both strong/weak acid had the same amount of H3O +.
But the amount of H3O+ in HCl is already the maximum, right from the start, since it's complete ionisation.
However, for CH3COOH, there were molecules of CH3COOH that remain undissociated. So during the titration, the H+ ions get used up, and the CH3COOH shifts to dissociate to form more H+ ions, hence a bigger volume of NaOH required for titration.

You jump out of thin air again!

Continue the thread. Repost the question: Assess the impacts on the environment of the developments in technology available to manufacture NaOH.

 

[danz90]

True. At the time they were initially measured, both strong/weak acid had the same amount of H3O +.
But the amount of H3O+ in HCl is already the maximum, right from the start, since it's complete ionisation.
However, for CH3COOH, there were molecules of CH3COOH that remain undissociated. So during the titration, the H+ ions get used up, and the CH3COOH shifts to dissociate to form more H+ ions, hence a bigger volume of NaOH required for titration.

How about any CH₃COO^- ions that undergo hydrolysis to produce OH^-?
That's why in weak acid strong base titrations the equivalence point is somewhere around 8-9.

 

[danz90]

So sorry that I hadnot seen this question until a few mins ago while surfin through the last few pages.

Supposing we have the same volume for both of them, obviously for acetic acid to have pH 3, it should be of much higher concentration than HCl due to its weaker ionization energy. While 10^-3 M is [HCl], with 1% molecules being ionized (somewhere around than, if I'm not wrong), acetic acid in this case needs to have a concentration of 0.1 M. And neutralizing depends on the concentration of the acid, not the pH, a bigger volume of the same alkaline solution is required.

Danz90's: Outline...

10 microgram/100mL= 100 microgram/1000mL=0.1 mg/L

- Weigh out exactly 1.6 g of lead nitrate
- Dissolve in a 1L of distilled water in a volumetric flask. (remember to make up to the exact calibration mark, avoid paralax, etc.)
- Using a clean 10mL pipette, transfer into another 1L volumetric flask, fill up to the mark, shake to ensure a uniform solution.
- Using a clean 10mL pipette, transfer 10 mL of this solution into another 1L volumetric flask, fill up and shake. Done!

Why is it 1.6g?

 

[Trebla]

PS: your question is evil.

Haha, I know it is . Well done for getting it out. I've got another "evil question" if you're up for it lol (though a lot less evil than the first one).

A galvanic cell is set up with two unknown electrodes (A and B) being used. Each solid electrode is immersed in their corresponding electrolytes solutions containing ions which are the same element as the electrode. (This means that for example, a Cu solid electrode is immersed Cu²⁺ solution)
A voltmeter is attached between the wires joining electrode A and electrode B. The reading on the voltmeter was 1.60 volts. Assuming ideal conditions what are the possible identities of electrode A and electrode B?

 

[SkimDawg]

A galvanic cell is set up with two unknown electrodes (A and B) being used. Each solid electrode is immersed in their corresponding electrolytes solutions containing ions which are the same element as the electrode. (This means that for example, a Cu solid electrode is immersed Cu²⁺ solution)
A voltmeter is attached between the wires joining electrode A and electrode B. The reading on the voltmeter was 1.60 volts. Assuming ideal conditions what are the possible identities of electrode A and electrode B?

That is just plain cruelty lol, I was trying to figure out a combination for 10 minutes and just gave up, i think the closest i got was:
Ni (anode) + Cl (cathode) = 1.62V
and Fe(3+) (cathode) and Zn (anode) = 1.53V
cmon danz, post up a qu

 

[danz90]

cmon danz, post up a qu

lol, ok... here goes. a nice and easy one

Identify a catalyst for the Haber reaction and explain why and how it lowers the temperature required for the reaction.

 

[Trebla]

That is just plain cruelty lol, I was trying to figure out a combination for 10 minutes and just gave up, i think the closest i got was:
Ni (anode) + Cl (cathode) = 1.62V
and Fe(3+) (cathode) and Zn (anode) = 1.53V

Haha, you were close with the second one. The answer is Zn (anode) with -0.76V and Mg (cathode) with 2.36V.
lol ok I'll stop asking these evil questions now....

 

[SkimDawg]

lol, ok... here goes. a nice and easy one

Identify a catalyst for the Haber reaction and explain why and how it lowers the temperature required for the reaction.

I think this is the 4th qu i've answered on the Haber process, ill be ready for any question on it in the exam haha.
A porous iron catalyst made from iron oxide is the catalyst used (i think)
Under normal atmospheric conditions, the reaction basically does not occur (K is very small). The reaction, in fact, needs a catalyst. In the presence of a catalyst, it has been found that reaction is most efficient when subjected to high pressures - several hundred times atmospheric pressure - but that increasing the temperature in fact causes a decrease in ammonia production. As well, a catalyst is used in the Haber process to lower the activation energy so that the N2 bonds and H2 bonds may be broken more readily. Hence, a catalyst lowers the temperature required for the reaction.

next qu: I've ran out of questions to ask lol....

 

[danz90]

I think this is the 4th qu i've answered on the Haber process, ill be ready for any question on it in the exam haha.
A porous iron catalyst made from iron oxide is the catalyst used (i think)
Under normal atmospheric conditions, the reaction basically does not occur (K is very small). The reaction, in fact, needs a catalyst. In the presence of a catalyst, it has been found that reaction is most efficient when subjected to high pressures - several hundred times atmospheric pressure - but that increasing the temperature in fact causes a decrease in ammonia production. As well, a catalyst is used in the Haber process to lower the activation energy so that the N2 bonds and H2 bonds may be broken more readily. Hence, a catalyst lowers the temperature required for the reaction.

next qu: I've ran out of questions to ask lol....

just to elaborate on ur answer. in the exam, i would refer to the catalyst as a finely ground, heterogeneous mixture of Fe₃O₄ / Fe_(s) which provides a very high surface area upon which N₂ / H₂ collisions can occur. As a result, there is less of a requirement of heat energy to provide the kinetic energy for particle collisions. Hence, the same rate of reaction can be achieved at lower temperatures with a catalyst. The catalyst itself does not increase the rate of reaction, but provides an alternate chemical pathway through which the SAME rate of reaction can be achieved at lower temperatures.

heres another easy one:

A number of Mt Isa children were found with blood lead levels above the recommended maximum of 10µg/100mL.

(a) Express the maximum recommended level of lead in ppm.

(b) Identify a reagent to detect lead ions in solution. Construct an ionic equation for this reaction.

(c) Explain why AAS is a more suitable method for analysing lead levels in blood, soil and surface water.

P.S. I'm going to go study Monitoring and Management. Will be back around 12.30

 

[danz90]

By the way, out of curiosity, who here did the Independent Trial for chem this year? How did u find it, what mark did u get?

 

[SkimDawg]

was just about to say that, I did this paper (that question you posted is from it ), I found it alright, the airbag question was gay though. I think i got ~80%, which reflected on the amount of though study I did (not much haha).

 

[danz90]

was just about to say that, I did this paper (that question you posted is from it ), I found it alright, the airbag question was gay though. I think i got ~80%, which reflected on the amount of though study I did (not much haha).

yehh i did the independent paper too. was a fairly difficult paper, I definitely agree. The airbag question was tricky to comprehend.. but once you wrote down a proper equation, the mole calculations were straightforward.

I got 86%.. which wasn't bad.. and hopefully will get at least that kind of raw mark in HSC Exam.

You going for a state ranking in chem?

 

[SkimDawg]

Well ill try and answer those questions anyway, as i got them wrong in the trial, because we hadn't even STARTED the topic in class /siighghghghh.
a)10ppm
b) I Found some nuts reagent for it, but ill just do the easy one. Add dilute HCl. A white precipitate forms which dissolves in hot water:
Pb(2+)aq + 2Cl(-)aq ---> PbCl(2)aq
Then, with a different sample of course, add KI and a yellow precipitate of PbI(2) forms:
Pb(2+) + 2I(-) ----> PbI(2)
The other reagent is called dithizone in 1,1,1 trichloroethane. Is this the one we are supposed to know? Theres no way I could remember that lol.
c) AAS is a more suitable method for analysing lead levels in blood, soil and surface water as the elements are in such a small amount that the analytical methods the chemists used before AAS were not sensitive enough to detect the low concentrations of these elements and so their presence had gone unnoticed. These elements are also known as trace elements in organisms (blahblah blah..). I'm getting to tired atm so i cant really elaborate on this, so im pretty much addressing only half of the question.