Chemistry 3/4 2013 Thread

[clıppy]

Insight 2013
Q21 - 20mL of distilled water is added to a 0.1 M ethanoic acid solution. As a result of this dilution, the
A - pH decreases and the percentage ionisation of the acid increases
B - pH drops and the percentage ionisation of the acid decreases
C - pH increases and the percentage ionisation decreases
D - pH does not change because ethanoic acid is a weak acid

The answers say A because

When water is added, the acid is diluted so the pH rises. However, the addition of water also pushes the reaction in the forward direction. This means the percentage ionisation increases.

However I chose B (knowing that percentage ionisation increased) because, if we added water in this equilibrium reaction the pH would decrease as more H+ ions are formed?

[eddybaha]

the amount of H+ ions increase but the concentration decreases. therefore ph increases.

[ahat]

If for question we found that I = 0.23 A (to 2. sig. figs) but the unrounded answer was I = 0.2345A, which answer would we use if the next part of the question calculate the energy? The rounded or exact answer? (because in the solutions, they've used the rounded one and I've always used exact).

[lzxnl]

Insight 2013
Q21 - 20mL of distilled water is added to a 0.1 M ethanoic acid solution. As a result of this dilution, the
A - pH decreases and the percentage ionisation of the acid increases
B - pH drops and the percentage ionisation of the acid decreases
C - pH increases and the percentage ionisation decreases
D - pH does not change because ethanoic acid is a weak acid

The answers say A because However I chose B (knowing that percentage ionisation increased) because, if we added water in this equilibrium reaction the pH would decrease as more H+ ions are formed?

Easiest way of looking at this: you're diluting an acid solution. There is no way you can increase any concentration of any solute.

[AwayBroadcast]

Might as well also ask, do we get consequential marks in chem? As in we don't lose answer marks for consequent parts of the question?

We do so if worse comes to worse and you can't figure out the answer to a question, but you need that answer for the following question or one later on, just write down a random number as the answer and continue on as if you are the most confident person in the world that that answer is correct!! haha

[eddybaha]

If for question we found that I = 0.23 A (to 2. sig. figs) but the unrounded answer was I = 0.2345A, which answer would we use if the next part of the question calculate the energy? The rounded or exact answer? (because in the solutions, they've used the rounded one and I've always used exact).

you use the exact value in your calc, but they usually accept a range of values anyways.

[Mafioso]

I get confused when calculating the mol of electrons. Say I am dealing with Al3+, when do I multiply the mol of electrons by 3 and when do I divide by 3?

[Edward21]

Insight 2013
Q21 - 20mL of distilled water is added to a 0.1 M ethanoic acid solution. As a result of this dilution, the
A - pH decreases and the percentage ionisation of the acid increases
B - pH drops and the percentage ionisation of the acid decreases
C - pH increases and the percentage ionisation decreases
D - pH does not change because ethanoic acid is a weak acid

The answers say A because However I chose B (knowing that percentage ionisation increased) because, if we added water in this equilibrium reaction the pH would decrease as more H+ ions are formed?

Think of our actions as the be all end all. Like if we dilute a solution and lower the concentration of all ionic species, the system will partially oppose this, but in the end we win It's an easy way to remember why the pH of an equilibrium or reversible reaction will never come back to what it wants to, because we win in the end, our change will be the lasting one no matter what the system tries to do.

[brightsky]

you really need to think about this logically and consider every situation separately. usually however, you can work it out using the molar ratios by using the following method:

2A + 5B --> 3C

if you work out n(A) = x mol, you write down the ratio:
n(A): n(B) = 2 : 5
and then divide it through by a number so that the left hand side ratio becomes 1.
n(A) : n(B) = 2 : 5 = 1 : 5/2
now multiply the number on the right hand side by n(A) to find n(B)
n(B) = 5/2 * n(A) = 5/2*x

now if you get something like MgCl2 and you found n(Cl) and were asked to find n(MgCl2) then you need to be careful. normally you would think Cl has a 2 after it so surely multiply by 2! no. n(Cl) remains unchanged. think about it this way. you have a fixed amount of Cl, and you want to distribute the Cl to boxes labelled MgCl2. each box has a capacity to hold 2 Cl atoms. so how many boxes are there? yes, 1/2 *n(Cl) not 2*n(Cl).

hope this makes sense.

[ahat]

A students has 10mL of a solution of KOH which has a pH = 10. The solution is diluted with deionized water to make up a litre solution. The pH of the diluted solution will be?

I used the formula:











However, this is incorrect. If at the start, you had worked with [OH^-] =10^-4, the pH would be calculated at 8. Why the difference?

Thanks

[Jeggz]

A students has 10mL of a solution of KOH which has a pH = 10. The solution is diluted with deionized water to make up a litre solution. The pH of the diluted solution will be?

I used the formula:











However, this is incorrect. If at the start, you had worked with [OH^-] =10^-4, the pH would be calculated at 8. Why the difference?

Thanks

It's because KOH is a base and hence the [OH] = and the [H+]=

[ahat]

It's because KOH is a base and hence the [OH] = and the [H+]=

I always thought pH was equivalent to [H⁺] = 10^-pH? 

I've attached their solution btw.

[Patches]

How do you work this out (the attachment)? I get C, since surely even if [HOBr] increased by the amount [OH-] decreases by, the [H+] will still be higher since the change is only partially reversed. This would increase the value of the fraction, I thought.

The answers say D, yet surely adding H+ is going to change the value of Q. Is it just that they've confused equilibrium concentration and the equilibrium constant, or is my logic wrong?


Also, does the C-O-C in a glucose or other sugar count as an ether linkage?

Thanks

[achre]

Insight paper asks: Name one method or change, other than altering the primary structure, which can alter an enzyme's catalytic ability.
I answered competitive inhibition, which isn't on the s.d. for chem and feels like more of a bio answer, but the answers Insight gave were 'alterations to temperature or pH'. Aren't alterations to temperature/pH just ways of altering the primary structure?

[Dismounted]

How do you work this out (the attachment)? I get C, since surely even if [HOBr] increased by the amount [OH-] decreases by, the [H+] will still be higher since the change is only partially reversed. This would increase the value of the fraction, I thought.

Note that the question tells us "...after a long period of time..." - this is telling us the system is at equilibrium.

Since we haven't changed the temperature of the system, the value of K remains the same. Thus, the value of the concentration fraction is the same as before we added the "strong acid" (Q = K). We assume the system was at equilibrium before the acid was added.

Also, does the C-O-C in a glucose or other sugar count as an ether linkage?

Yes.