Stankovic123's chem q's

[zvezda]

If by 'normal' you mean it goes clear again, it's because you are really close to the equivalence point, but you haven't passed it yet to get to the endpoint where the colour change is permanent. 

I don't think change in colour of the aliquot has any correlation with amount of solution from burette.

I appreciate the help, but im looking for a more in depth explanation. Like, the colour change occurs because the solution in the burette has reacted with the one in the aliquot yeah? So why does this colour then disappear? Its not like the products have reacted (back reaction?)

[Limista]

ohh right. Yeah that's a good question, which I'm not sure I can provide a suitable answer to. However, I think it's got something to do with equilibrium? Maybe it's also got something to do with incomplete ionisation of the acid, and maybe this reversal of colour-change is more noticeable if the acid is weak?

[lzxnl]

It's to do with the reaction rate. I'm going to assume that you're titrating with acid. As you add more acid and as the amount of base decreases, the concentration of base drops. Therefore, the reaction rate begins to slow down and the acid colour lingers for longer. Eventually, it lingers permanently when you reach the equivalence point.

[Mao]

Was just thinking about titrations (as you do), and was windering about the immediate colour change in the aliquot before the colour returns back to normal. Why does this happen?
Shouldnt the change in colour of the aliquot be proportional to the amount of solution added from the burette?
Thanks

There are actually three things happening.

For a titration in a controlled environment, it's a combination of mixing and reaction rates. In an uncontrolled environment, atmospheric contaminants will also alter the solution.

Before I get into that, let's look at the physical changes. For now, consider an acidic aliquot, a basic titrant and phenolphthalein indicator.
- The indicator (which are molecules like everything else) is evenly dispersed in the aliquot.
- A drop of titrant (basic, concentrated) is added to the aliquot (acidic, dilute). This forms a temporary region where the concentration of base is higher than acid.
- The indicator molecules in this region will react with base and become pink.
- As mixing takes place, the temporary region disperses. The concentration of base becomes even throughout the solution, and most of the base is neutralised by reaction with acid.
- As mixing/reaction takes place, deprotonated indicator (pink, basic) must find extra H+ in the solution to become clear again.

So, when you add a drop of base into acid, a small region of the liquid turns pink, but then quickly disappear. This is because the drop of base does not immediately mix with the liquid, so a small region of the liquid is more basic, and the indicator in that region takes on the purple basic form. This quickly disappear as the solution mixes and all the base are neutralised (indicator becomes clear again).

As you approach the equivalence point, the concentration of acid in the aliquot becomes very small. This drastically decrease the reaction rate. Even after adequate mixing has occurred, the solution can still remain pink, as the acid/base eventually neutralise each other, and the deprotonated indicator molecules (pink, basic) eventually find spare H+ to become protonated (clear, acidic).  The temporary pink colour can remain for a few seconds at most.

However, a weird observation is that even after you've reached (or slightly passed) the equivalence point, sometimes the solution can go back to clear after 10 or 15 minutes. This is contrary to the belief that once you've reached the end point, it's 'permanent'. This is caused by atmospheric CO2 acidifying your nearly-neutral solution, via CO2(g) + 2H2O(l) --> HCO3-(aq) + H3O+(aq). This is what I mean by 'controlled' vs 'uncontrolled' environments: a totally controlled environment would have a totally inert atmosphere, such as pure N2.

Neat (pure, deionised) water equilibrated in normal atmosphere has a pH of 5.4~5.7. Any neutral/basic solution would tend towards this pH if left in open atmosphere for long enough: enough CO2 will eventually dissolve in the solution to turn it slightly acidic again. Because of CO2, we still don't quite understand the precise origin of water's surface tension and its surface composition. It's a bitch, really

[zvezda]

There are actually three things happening.

For a titration in a controlled environment, it's a combination of mixing and reaction rates. In an uncontrolled environment, atmospheric contaminants will also alter the solution.

Before I get into that, let's look at the physical changes. For now, consider an acidic aliquot, a basic titrant and phenolphthalein indicator.
- The indicator (which are molecules like everything else) is evenly dispersed in the aliquot.
- A drop of titrant (basic, concentrated) is added to the aliquot (acidic, dilute). This forms a temporary region where the concentration of base is higher than acid.
- The indicator molecules in this region will react with base and become pink.
- As mixing takes place, the temporary region disperses. The concentration of base becomes even throughout the solution, and most of the base is neutralised by reaction with acid.
- As mixing/reaction takes place, deprotonated indicator (pink, basic) must find extra H+ in the solution to become clear again.

So, when you add a drop of base into acid, a small region of the liquid turns pink, but then quickly disappear. This is because the drop of base does not immediately mix with the liquid, so a small region of the liquid is more basic, and the indicator in that region takes on the purple basic form. This quickly disappear as the solution mixes and all the base are neutralised (indicator becomes clear again).

As you approach the equivalence point, the concentration of acid in the aliquot becomes very small. This drastically decrease the reaction rate. Even after adequate mixing has occurred, the solution can still remain pink, as the acid/base eventually neutralise each other, and the deprotonated indicator molecules (pink, basic) eventually find spare H+ to become protonated (clear, acidic).  The temporary pink colour can remain for a few seconds at most.

However, a weird observation is that even after you've reached (or slightly passed) the equivalence point, sometimes the solution can go back to clear after 10 or 15 minutes. This is contrary to the belief that once you've reached the end point, it's 'permanent'. This is caused by atmospheric CO2 acidifying your nearly-neutral solution, via CO2(g) + 2H2O(l) --> HCO3-(aq) + H3O+(aq). This is what I mean by 'controlled' vs 'uncontrolled' environments: a totally controlled environment would have a totally inert atmosphere, such as pure N2.

Neat (pure, deionised) water equilibrated in normal atmosphere has a pH of 5.4~5.7. Any neutral/basic solution would tend towards this pH if left in open atmosphere for long enough: enough CO2 will eventually dissolve in the solution to turn it slightly acidic again. Because of CO2, we still don't quite understand the precise origin of water's surface tension and its surface composition. It's a bitch, really

Nice and thorough. Appreciate it heaps. Thanks Mao

[zvezda]

Hey,
For any given equilibrium reaction, wouldnt the amount of starting material affect the K constant? So technically, despite being the same reaction, K can differ depending on the mol of reactants initially cant it?

[lzxnl]

Hey,
For any given equilibrium reaction, wouldnt the amount of starting material affect the K constant? So technically, despite being the same reaction, K can differ depending on the mol of reactants initially cant it?

No. The equilibrium constant is a function of the concentrations such that at these concentrations, there is no potential to keep reacting. If you like, the reaction is at the bottom of a hypothetical hill if the reaction quotient equals the equilibrium constant. It won't go in either direction. The reaction quotient expression also takes into account different combinations of reactants and products; the amount of starting material doesn't affect the constant.

[zvezda]

No. The equilibrium constant is a function of the concentrations such that at these concentrations, there is no potential to keep reacting. If you like, the reaction is at the bottom of a hypothetical hill if the reaction quotient equals the equilibrium constant. It won't go in either direction. The reaction quotient expression also takes into account different combinations of reactants and products; the amount of starting material doesn't affect the constant.

So even if i had 2 separate reactions of the haber process:
1- start with 2 mol of nitrogen and hydrogen
2- start with 2.5 mol of nitrogen and 2 mol hydrogen

The K values will be equal?

Oh i see it now.
Thanks

[brightsky]

yeah the concentration of starting material determines the concentration fraction/reaction quotient at time t = 0. the equilibrium constant of a reaction is fixed (unless there is a temp change). the system will move towards equilibrium if the concentration fraction does not equal the equilibrium constant.

[zvezda]

Another question:
Say you had an equilibrium reaction with the reactants and products in a gaseous state.
If the concentration of the products was higher than the concentration of the reactants, yet the number of molecules per mol of reaction is equal on both sides of the equation, wouldnt a change in volume/pressure cause a change to the equilibrium position?
I cant see how it wouldnt at the moment. If for example one decreased the volume of the system, wouldnt the rate of the back reaction increase further than the forward one given that the concentration of the products is higher than that of the reactants?

[lzxnl]

Another question:
Say you had an equilibrium reaction with the reactants and products in a gaseous state.
If the concentration of the products was higher than the concentration of the reactants, yet the number of molecules per mol of reaction is equal on both sides of the equation, wouldnt a change in volume/pressure cause a change to the equilibrium position?
I cant see how it wouldnt at the moment. If for example one decreased the volume of the system, wouldnt the rate of the back reaction increase further than the forward one given that the concentration of the products is higher than that of the reactants?

Wat? Dat made no sense watsoeva

If the concentration of the products was higher than that of the reactants, that just means K>1, especially if the number of molecules per mole of reaction is equal on both sides of reaction.
A change in volume/pressure would change absolutely nothing in this case. By changing the volume or pressure, you don't add or subtract a fixed number to the concentrations; you multiply them by the same number. As the number of molecules is the same on either side of the reaction, the reaction quotient would be constant.

[zvezda]

Wat? Dat made no sense watsoeva

If the concentration of the products was higher than that of the reactants, that just means K>1, especially if the number of molecules per mole of reaction is equal on both sides of reaction.
A change in volume/pressure would change absolutely nothing in this case. By changing the volume or pressure, you don't add or subtract a fixed number to the concentrations; you multiply them by the same number. As the number of molecules is the same on either side of the reaction, the reaction quotient would be constant.

Lol. Ill try rephrase.
If you had different concentrations on both sides of the reaction, wouldnt a change in volume have a more significant influence for the side with a higher concentration?

You know what, maybe i should just ask what other factors influence the rates of the forward and back reactions in a system, because despite having different concentration levels on both sides of the reaction, a system can still be in equilibrium.

Apologies for the rubbish phrasing

[brightsky]

yeah I sort of see where you are coming from.

consider the haber process: N2(g) + 3H2(g) <---> 2NH3(g), ΔH = -92.4 kJ/mol. suppose you chuck random quantities of N2, H2 and NH3 into a vessel and allow the system to reach equilibrium (e1). Kc = [NH3]e1^2/([N2]e1[H2]e1^3). importantly. all gas concentrations would be constant.

now suppose you decreased the volume. all gas concentrations would increase BY THE SAME FACTOR (and since the LHS has more gas particles, the LHS of the equation will benefit more). Qc = x^2[NH3]e1^2/x([N2]e1x^3[H2]e1^3) = 1/x^2 Kc, where x > 1. clearly, Qc < Kc. so the equilibrium position will shift to the right. as far as rates are concerned, at the moment the volume is decreased, both the forward and the back reaction will shoot upwards. however, the forward reaction will shoot up more because there are more gas particles on the LHS. afterwards, the forward reaction will decrease as the LHS species get consumed, and the back reaction will increase as more of the RHS species are produced, until the rates are again equal.

most imposed changes will influence the rate of both the forward and the back reaction in some way. exceptions include: changing the pressure of the vessel at constant temp when all species are aqueous, adding an inert gas at constant temp and volume, etc.

[lzxnl]

If you double the volume, then ALL concentrations are halved. It doesn't matter which species was originally at a higher concentration; its concentration is now halved. If you halve all the concentrations in a reaction quotient where the sum of the indices of the top and bottom are the same, you're not going to affect the value of the reaction quotient.

[zvezda]

yeah I sort of see where you are coming from.

consider the haber process: N2(g) + 3H2(g) <---> 2NH3(g), ΔH = -92.4 kJ/mol. suppose you chuck random quantities of N2, H2 and NH3 into a vessel and allow the system to reach equilibrium (e1). Kc = [NH3]e1^2/([N2]e1[H2]e1^3). importantly. all gas concentrations would be constant.

now suppose you decreased the volume. all gas concentrations would increase BY THE SAME FACTOR (and since the LHS has more gas particles, the LHS of the equation will benefit more). Qc = x^2[NH3]e1^2/x([N2]e1x^3[H2]e1^3) = 1/x^2 Kc, where x > 1. clearly, Qc < Kc. so the equilibrium position will shift to the right. as far as rates are concerned, at the moment the volume is decreased, both the forward and the back reaction will shoot upwards. however, the forward reaction will shoot up more because there are more gas particles on the LHS. afterwards, the forward reaction will decrease as the LHS species get consumed, and the back reaction will increase as more of the RHS species are produced, until the rates are again equal.

most imposed changes will influence the rate of both the forward and the back reaction in some way. exceptions include: changing the pressure of the vessel at constant temp when all species are aqueous, adding an inert gas at constant temp and volume, etc.

Yeah i suppose i can see how its works through the mathemaics of it all, but not so much in a practical sense. The concentrations of reactants and products in whatever equilibrium system, as you guys said, change by the same factor. So then, why does the equilibrium position change? Because the frequency of successful collisions for both back and forward reactions should then be changed by the same factor and remain the same if there was a change in volume no?