Chemistry Predictions/Thoughts (4 Viewers)

[anonymoushehe]

so temp increases right

No the partial pressures of the species in the equilibrium is constant but the total pressure increases - the equilibrium will only shift if the pressure of the species actually changes which hasn’t here

 

[hm...mmmm]

Wait so saying 2 precipitates forming is wrong? I did think of the assumption that mg goes to completion but idk i calculated both

 

[YulianaTheAmogusArtist]

can someone explain 19 please

 

[hm...mmmm]

Do you guys think a raw 78 will allign upto 84ish?

 

[dubbatee]

can someone explain 19 please

So the first part of the question is the rxn between the acetate and the HCl. We see that we actually neutralise exactly half of the acetate, so we have 0.5 mols of acetate and 0.5 mols acetic acid. ie equimolar amounts of weak acid and its weak conj. base (buffer). You can calculate the Ka of acetic acid using an ICE table, but since the concentration of acetate and acetic acid are so close, we know that pH = pKa. Now that we know the Ka of the acid, we can use this to construct another ICE table, now using the fact that our solution is 1L in volume. Our initial conditions are 0.05M for acetate and acetic acid (easier to consider the system prior to any shifting) and 0M for the H+. You can then calculate the shift since we know the Ka and thus we can find the pH of the solution, but since our shift will be very very small, we recognise that the concentration of acetic acid and acetate will be very similar, so once again [H+] = Ka, and thus pH = pKa, ie pH = 4.8.

It may be easier to see this when you write out the equilibrium expression

 

[Ejitt]

View attachment 50270
No temperature change

aint no way we trusting chat on this.
Addition of argon increases pressure, causes eq to shift right, as delta H<0 for fwd reaction, T increases. Reason it’s 4 not 3 is cos of added difficulty realising what argon does as apposed to 3 markers, which would just say: pressure increased, what happens to T?

 

[anonymoushehe]

aint no way we trusting chat on this.
Addition of argon increases pressure, causes eq to shift right, as delta H<0 for fwd reaction, T increases. Reason it’s 4 not 3 is cos of added difficulty realising what argon does as apposed to 3 markers, which would just say: pressure increased, what happens to T?

The change in pressure has to be from the species in the actual equilibrium not from an inert has

 

[dubbatee]

aint no way we trusting chat on this.
Addition of argon increases pressure, causes eq to shift right, as delta H<0 for fwd reaction, T increases. Reason it’s 4 not 3 is cos of added difficulty realising what argon does as apposed to 3 markers, which would just say: pressure increased, what happens to T?

Whilst the overall pressure of the system is affected, the partial pressures (the pressure that each individual gas is exerting) does not change. The change in partial pressure is what actually causes our system to shift when we deal with pressure changes. Since the partial pressures for the two gases in the equilibrium are unchanged, the equilibrium position is not disturbed, so no temperature change should be observed

 

[YulianaTheAmogusArtist]

So the first part of the question is the rxn between the acetate and the HCl. We see that we actually neutralise exactly half of the acetate, so we have 0.5 mols of acetate and 0.5 mols acetic acid. ie equimolar amounts of weak acid and its weak conj. base (buffer). You can calculate the Ka of acetic acid using an ICE table, but since the concentration of acetate and acetic acid are so close, we know that pH = pKa. Now that we know the Ka of the acid, we can use this to construct another ICE table, now using the fact that our solution is 1L in volume. Our initial conditions are 0.05M for acetate and acetic acid (easier to consider the system prior to any shifting) and 0M for the H+. You can then calculate the shift since we know the Ka and thus we can find the pH of the solution, but since our shift will be very very small, we recognise that the concentration of acetic acid and acetate will be very similar, so once again [H+] = Ka, and thus pH = pKa, ie pH = 4.8.

It may be easier to see this when you write out the equilibrium expression

i didnt realise its a buffer, Im such a fool. THANKSS though for the explanation, very helpful.

 

[RedDragonStar1]

ok judging by the discussion so far, i can conclude that i clearly got cooked by this exam (hopefully still at least b5)

 

[dubbatee]

i didnt realise its a buffer, Im such a fool. THANKSS though for the explanation, very helpful.

Dont worry too much, these questions that involve recognising that both the weak acid and conjugate base still exist in solution after the reaction are notoriously difficult because during the exam you are more stressed etc. I think only the very top of the chemistry cohort will get this question, which is why its Q.19, its meant to be very very hard.

 

[nowomansland]

guys would we lose a mark if we didn’t give name for the last q it was only 4 marks and the q didn’t ask for the name it just said structure

 

[HappyTheSkibidi]

guys would we lose a mark if we didn’t give name for the last q it was only 4 marks and the q didn’t ask for the name it just said structure

Nah don't think so, as long as you drew it out or gave some indication of what it was,
Is it just me or was this paper genuinely one of the easiest one yet?

 

[almost_done]

 

[tsswhzl]

Does anyone have all the Mc answers

 

[HappyTheSkibidi]

View attachment 50273

It doesn't change KEQ because KEQ is only changed by Temperature.
In the system however it is still dynamic equilibrium, by adding Argon it increases the pressure of the system and causes it to shift towards the right.

 

[thatnerdyhylian]

This argon question looks like the hardest in the paper by the looks of this conversation

 

[dubbatee]

We can probably figure them out together right? I think it's the following, could be wrong

BBAACDDDADABCBDAAACB

Q.18 is a question based on Mohr's Precipiation titration method, but its kind of worded strangely. Normally, we add Ag+ (unknown) into the yellow solution (has potassium dichromate) with some known concentration of Cl-, and the Ag+ ions should react with the Cl- ions in the solution. After essentially all the Cl- ions have been reacted, the Ag+ will react with the dichromate ions, turning the solution red. The colour should go from yellow to red, and the AgCr2O4 has a slightly greater molar solubility.

But its worded sort of poorly, I think I would put B though. Theres no way they have added the Silver directly with the indicator initally, because then this titration cant actually measure anything. So i think the known AgNO3 solution is in the burette, and the unknown Cl- solution is in the conical flask with the indicator. We reach endpoint when the silver has reacted with all the Cl- ions, and then it finally reacts with the dichromate, making the soln turn red.

 

[thatnerdyhylian]

So i think the known AgNO3 solution is in the burette, and the unknown Cl- solution is in the conical flask with the indicator.

You're determining the concentration of silver ions though, aren't you?

 

[thatnerdyhylian]

Oh that's my bad, I understand what you mean