brightsky's Chem Thread

[brightsky]

from the same exam multi-choice q27.

Aniline C6H5NH2 is a weak base and ionises in water as shown in the equation below:
C6H5NH2(aq) + H2O(l) <--> C6H5NH3+(aq) + OH-(aq)

If a solution of aniline is diluted with water, what will happen to the pH and the percentage ionisation?

okay so dilution causes the position of equilibrium to shift to the right, since we want more particles to counteract with increase in volume. so % ionisation obviously increases.

but [OH-] will go up. surely this results in pH INCREASE. the answers, however, say that it is a pH decrease...I cannot see how this is right.

[brightsky]

also, when vcaa gives pH values, do they need to be taken into account when determining sig figs of final answer? say pH = 6.9. then [H+] = 10^-6.9 = 1*10^(-7) which is a little restrictive but correct....what if they give pH = 6? how are we supposed to express answer correct to 0 sig figs?

[SocialRhubarb]

how are we supposed to express answer correct to 0 sig figs?

[lzxnl]

from the same exam multi-choice q27.

Aniline C6H5NH2 is a weak base and ionises in water as shown in the equation below:
C6H5NH2(aq) + H2O(l) <--> C6H5NH3+(aq) + OH-(aq)

If a solution of aniline is diluted with water, what will happen to the pH and the percentage ionisation?

okay so dilution causes the position of equilibrium to shift to the right, since we want more particles to counteract with increase in volume. so % ionisation obviously increases.

but [OH-] will go up. surely this results in pH INCREASE. the answers, however, say that it is a pH decrease...I cannot see how this is right.

You've made the same mistake as before somewhere.

I see, thankyou!


Also, when water is added to a dilute solution of NH3(aq), does the pH of the solution rise and fall?

NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq), so if [H2O] increases, then net forward reaction occurs, increasing [OH-] and so raising the pH, no? The solution key suggests otherwise...where have I gone wrong?

Thanks!

brightsky, use some common sense here. If you dilute an ammonia solution, there's no way you're going to increase the hydroxide ion concentration. Common sense trumps LCP any day.

NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)

OK, so you add some water. What's the first thing that happens?  The hydroxide ion concentration decreases. Almost instantly. LCP says that a system can only PARTIALLY counteract changes. Here, the dilution has reduced the number of particles per unit volume. All the system can do is partially increase the number of particles per unit volume with a net forward reaction. However, this is only partially opposing. It can't ever recover the change in [OH-] by itself.

Another way of thinking about this:
K=[NH4+][OH-]/[NH3] at equilibrium which is fixed.
Introduce more water. Let's assume you halve the concentrations.
So your new Q value is half of K. The system will try to increase Q to K.
What has changed? The concentration of ammonia has decreased too from the dilution. Due to the mole ratios, concentrations of ammonium ion and hydroxide ion are going to be the same. Now, if [NH3] is lower, then the concentrations of ammonium ion and hydroxide ion at equilibrium cannot possibly be equal to their former values. Previously, [NH3] was higher, so [NH4+] and [OH-] could also be larger. But now that [NH3] is smaller, [NH4+] and [OH-] must be smaller to allow Q to equal K. So despite the fact that the system undergoes a net forward reaction, that just means that after adding the water, your Q is lower than K, so the reaction proceeds forward. The number of MOLES of OH- has increased; the concentration hasn't.

Now replace ammonia with aniline

To cut a long story short
You're diluting the hydroxide ion concentration. The system cannot fully counter that changes. pH decreases.

also, when vcaa gives pH values, do they need to be taken into account when determining sig figs of final answer? say pH = 6.9. then [H+] = 10^-6.9 = 1*10^(-7) which is a little restrictive but correct....what if they give pH = 6? how are we supposed to express answer correct to 0 sig figs?

So...undefined is now correct to zero sig figs?

[drake]

from the same exam multi-choice q27.

Aniline C6H5NH2 is a weak base and ionises in water as shown in the equation below:
C6H5NH2(aq) + H2O(l) <--> C6H5NH3+(aq) + OH-(aq)

If a solution of aniline is diluted with water, what will happen to the pH and the percentage ionisation?

okay so dilution causes the position of equilibrium to shift to the right, since we want more particles to counteract with increase in volume. so % ionisation obviously increases.

but [OH-] will go up. surely this results in pH INCREASE. the answers, however, say that it is a pH decrease...I cannot see how this is right.

lol brightsky, we make the same mistake every time haha

[SocialRhubarb]

So...undefined is now correct to zero sig figs?

Haha, how would you represent zero sig figs? : )

[brightsky]

You've made the same mistake as before somewhere.

lol brightsky, we make the same mistake every time haha

omg shoot me...

thanks nliu! can you have a look at the other questions?

[lzxnl]

Haha, how would you represent zero sig figs? : )

Erm...zero sig figs, by virtue of the name, suggests that none of the digits is known. So...perhaps pH 6 under that convention would really mean pH from 5.5-6.5?
Or we could interpret it to be infinitely inaccurate

[Mao]

also, when vcaa gives pH values, do they need to be taken into account when determining sig figs of final answer? say pH = 6.9. then [H+] = 10^-6.9 = 1*10^(-7) which is a little restrictive but correct....what if they give pH = 6? how are we supposed to express answer correct to 0 sig figs?

Sig figs is not a very precise way of dealing with uncertainties. It is at best an estimate of the uncertainty to the nearest order of magnitude. When in doubt, always apply common sense and logic before locking yourself up in sig fig rules.

e.g. if we are reading the pH from a pH metre, which only shows 6.9, then the true pH is somewhere between 6.85 and 6.95. So . The only figure we are sure about is the first one, so the answer to the correct significant figures is .

If the pH shows only 6 and no digits, then the true pH is somewhere in between 5.5 to 6.5, so . As you can see here, we are not sure what the first digit should be. A pH of 6 with no further precision represents an uncertainty range of an entire decade, and so we can never be sure what the leading number is in a decimal number system. In this case, there is no good way to write down the answer using the significant figures convention. The best you can do is to indicate the answer is within the decade of .

Typically, we just leave everything in the log scale, to avoid this asymmetric mapping of uncertainties from the logarithmic scale to the linear scale.

TL;DR significant figures is quick, but not always applicable.

[brightsky]

i see...I see...hopefully vcaa doesn't do something dodgy...

with regard to proton NMR, for a question that gives the number of peaks, the area under each peak, the splitting pattern and the chemical shift of each peak, and asks that we identify the type of proton in each environment, does it suffice to simply draw what is given in the data book? for instance, CH3 - R or something. or do we need to be more specific? STAV puts: "quartet at 2.0 is a -CH2 group split by an adjacent - CH3 and bonded to -CO-". is this necessary?

thanks!

[brightsky]

STAV 2011 MC Q14. Methyl orange has a pH range of 3.1 - 4.4. The equivalence point is at pH = ~5.5. The graph sort of curves away after pH = 5. Surely methyl orange would not be appropriate? STAV says that the answer is C, however.

[lzxnl]

STAV 2011 MC Q14. Methyl orange has a pH range of 3.1 - 4.4. The equivalence point is at pH = ~5.5. The graph sort of curves away after pH = 5. Surely methyl orange would not be appropriate? STAV says that the answer is C, however.

What are the other choices? And what's the reaction? Bear in mind that adding half a drop may be the difference between pH 5.5 and pH 4. You've seen how steep titration curves can be.

[brightsky]

okay here's a crappy screenshot of the question.

the curve does plummet around the middle, but even then, methyl orange seems to me a pretty bad choice of indicator; the curve 'curves' away at around pH 3 - 4. surely this will affect the accuracy of the concentration calculated to more than a negligible extent.

also, nliu, can you have a look at the NMR question above? cheers.

[lzxnl]

i see...I see...hopefully vcaa doesn't do something dodgy...

with regard to proton NMR, for a question that gives the number of peaks, the area under each peak, the splitting pattern and the chemical shift of each peak, and asks that we identify the type of proton in each environment, does it suffice to simply draw what is given in the data book? for instance, CH3 - R or something. or do we need to be more specific? STAV puts: "quartet at 2.0 is a -CH2 group split by an adjacent - CH3 and bonded to -CO-". is this necessary?

thanks!

I'm not actually sure, but to be safe, I'd make it absolutely clear which protons I was referring to.

STAV 2011 MC Q14. Methyl orange has a pH range of 3.1 - 4.4. The equivalence point is at pH = ~5.5. The graph sort of curves away after pH = 5. Surely methyl orange would not be appropriate? STAV says that the answer is C, however.

If you look carefully, pH 4.4 is about half a drop away from the equivalence point. That's close enough.

[brightsky]

I'm not actually sure, but to be safe, I'd make it absolutely clear which protons I was referring to.

If you look carefully, pH 4.4 is about half a drop away from the equivalence point. That's close enough.

diu the subjectivity...

okay thanks nliu!