Chemistry help (1 Viewer)

[A.void.that.thinks]

Just need some help with the following questions

1) I chose A but the answer is B and I don’t know why.


2) I chose C but the answer is D



3) Also, with indicators, how come they can only be made out of weak acids/bases. Why not a strong acid/base?

 

[username_2]

Just need some help with the following questions

1) I chose A but the answer is B and I don’t know why.
View attachment 32838

2) I chose C but the answer is D
View attachment 32839


3) Also, with indicators, how come they can only be made out of weak acids/bases. Why not a strong acid/base?

So the idea for question 15 is to identify that the reactants shouldn't be completely reacted off to create an equilibrium (as only water will be left, which is not a buffer). With this, consider the eqn:
CH3COOH(aq) + OH−(aq) → CH3COO−(aq) + H2O(l)

Mole ratio: 1:1 to completely react each other off. (Notice that if A happens, only CH3COO- and H2O will be left, which doesn't create a buffer.
So A is eliminated, similarly D is eliminated. If there's more moles of NaOH, then it will completely react off, just resulting in excess of NaOH being left. Hence, C is eliminated. Therefore, B is the only viable option as the reaction results in some CH3COOH being left and some CH3COO-, which does create a buffer. This could be put in better words but that's the idea. Hope it helps.

Edit: Note for indicators that they must be weak acid/base systems because that is what creates a buffer system (i.e. a system that is in equilibrium). As mentioned, if a strong base and acid react, they will just produce water only and nothing else, which obviously doesn't indicate the acidity or basicity of a substance.

 

[CM_Tutor]

Yes, @Hivaclibtibcharkwa, you need a mixture of a weak acid or base and its conjugate. So, yes, you can't have NaOH as one of the two components in a buffer.

However, as @username_2 is explaining, you can use NaOH to make one of those components.

If I have 0.50 mol CH₃COOH and I add 0.25 mol NaOH, think of a limiting reagent problem. I will end up with 0.25 mol of CH₃CO₂Na, the sodium salt of acetic acid, and 0.25 mol of unreacted CH₃COOH as it is the reagent in excess. So, after all the NaOH is used up, I have a mixture with

[CH₃COOH] : [CH₃CO₂^-] = 1 : 1​

which is perfect for a buffer.

With (A), I would have 0.50 mol of CH₃CO₂Na in water, which is not a buffer.

With (C), I would have a 1 : 1 mixture of CH₃CO₂Na and NaOH, which is but a buffer.

With (D), I would have 0.25 mol of Ba(CH₃CO₂)₂, which is not a buffer.

 

[CM_Tutor]

2) I chose C but the answer is D
View attachment 32839

Hydrogen environments w and y are isolated as they have no H atoms on the adjacent atom, so will be singlets.

Hydrogen environment x has one neighhbouring environment (the one z hydrogen) and so will appear as 1 + 1 = 2 lines, a doublet.

Hydrogen environment z has one neighhbouring (the two x hydrogens) and so will appear as 2+ 1 = 3lines, a triplet.

Hydrogen environment y would be coupled with / have splitting from hydrogen environment z if the methyl group in y was bonded directly to the carbon involved with environment z, but it is not - there is an O atom between them, which isolates y and z from each other.

 

[CM_Tutor]

3) Also, with indicators, how come they can only be made out of weak acids/bases. Why not a strong acid/base?

For an acid-base indicator to work, it must have two forms that have different colours, and must be able to move from one form to the other just after the equivalence point. If one of the forms was a strong acid, then it would immediately ionise to the other form and only display one colour. So, an indicator must be weak as it needs to move from mostly HInd (its acid form) to Ind^- (its basic form) with changes in pH.