VCE Chemistry Question Thread

[keltingmeith]

Hi, I need some help with this question please

What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

I'm able to get the correct answer. Any help would be appreciated

The last time you posted this, I asked a question in reference to your working.

[RazzMeTazz]

Hi, I need some help with this question please

What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

I'm able to get the correct answer. Any help would be appreciated

What's the answer? Is it 0.776L?

[stewartmahi]

Hi guys need help with this question

[Redoxify]

I have the same question as stewart.mahi,
can someone please help me out with that question, it's due tomorrow

[Chazef]

Hi guys need help with this question

went to this site https://www.erowid.org/archive/rhodium/chemistry/equipment/ph-indicator.html and it looks like red beet goes from yellow to red as ph drops from 13 to 10 which tells me that the deprotonated beet i.e. Beet- is yellow and the protonated one i.e. HBeet is red. So say something like Beet-(aq) + H+(aq) --> HBeet(aq) and underneath state the colours of the beets. I think that's how you go about the question

[stockstamp]

Where'd you get these two steps from?

Sorry, I missed that reply^
Again, here is what I did

What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

• NaOH (aq) --> Na⁺ (aq) + OH^- (aq)
• KOH (aq) --> K⁺ (aq) + OH^- (aq)
• n(NaOH) = n(OH^-) = 0.5/40 = .0125 mol
• n(KOH) = n(OH^-) = 0.8/56.1 = 0.0142603 mol
• n(OH^-)total = 0.0125 + 0.0142603 = 0.027603 mol
• Therefore n(H₃O⁺) required to neutralise is also = 0.027603 mol
• H₂SO₄ (aq) --> 2H₃O⁺
• n(H₂SO₄) required to neutralise = 1/2 n(H₃O⁺) = 0.0138015
• V(H₂SO₄) = n/C = 0.0138015/0.1 = 0.138 Litres

You asked about these two steps:
1.    Therefore n(H₃O⁺) required to neutralise is also = 0.027603 mol
2.    H₂SO₄ (aq) --> 2H₃O⁺


1: I am assuming that because I have 0.027603 mol of OH^- (the ion that makes the solution basic that I would need the same number of mols of H₃O⁺ (the ion that makes it acidic) to neutralise the solution.

2: For every mol of H₂SO₄ placed in water, there are two moles of H₃O⁺.  This seems necessary when trying to work out how many moles of  H₂SO₄ are present if you're working off the assumption (which you are) that the solution was neutralised.

If you can't make sense of this could someone show their own method of finding an answer? Thanks

[keltingmeith]

1: I am assuming that because I have 0.027603 mol of OH^- (the ion that makes the solution basic that I would need the same number of mols of H₃O⁺ (the ion that makes it acidic) to neutralise the solution.

This is true, but you can't just assume it - why do you think this is the case?

2: For every mol of H₂SO₄ placed in water, there are two moles of H₃O⁺.  This seems necessary when trying to work out how many moles of  H₂SO₄ are present if you're working off the assumption (which you are) that the solution was neutralised.

This is true - but HOW do you know that it's true?

Anyway, I've had a look through your working a bit more thoroughly (sorry for not doing this earlier, I must seem like a dick. >.<), and I think you've just made an arithmetic error. Specifically, (1/2)*0.027603=0.01338015. Otherwise, it looks right to me.

[lzxnl]

This is true - but HOW do you know that it's true?

Technically it's not true. HSO₄^- is a weak acid with a K_a of around 0.01, so it's not actually fully deprotonated in water. A 1 M sulfuric acid solution has a proton concentration significantly closer to 1 M than it does to 2 M. The only reason why we deprotonated it twice is because it's reacting with a strong base (and because it's a much stronger acid than most other weak acids out there). When reacting with sodium hydroxide, nearly all acids become fully deprotonated (exceptions include the really weak ones out there like NH₄⁺ and ethanol).

[keltingmeith]

Technically it's not true. HSO₄[/sup]-[/sup] is a weak acid with a K_a of around 0.01, so it's not actually fully deprotonated in water. A 1 M sulfuric acid solution has a proton concentration significantly closer to 1 M than it does to 2 M. The only reason why we deprotonated it twice is because it's reacting with a strong base (and because it's a much stronger acid than most other weak acids out there). When reacting with sodium hydroxide, nearly all acids become fully deprotonated (exceptions include the really weak ones out there like NH₄⁺ and ethanol).

Urp, I read it as you need to lose two hydroniums to neutralise it. Mah bad. :S

[thushan]

Technically it's not true. HSO₄[/sup]-[/sup] is a weak acid with a K_a of around 0.01, so it's not actually fully deprotonated in water. A 1 M sulfuric acid solution has a proton concentration significantly closer to 1 M than it does to 2 M. The only reason why we deprotonated it twice is because it's reacting with a strong base (and because it's a much stronger acid than most other weak acids out there). When reacting with sodium hydroxide, nearly all acids become fully deprotonated (exceptions include the really weak ones out there like NH₄⁺ and ethanol).

Even ammonium will be completely deprotonated if you chuck in an excess of NaOH as opposed to a stoichiometrically equal amount.

[Eiffel]

if hydrochloric acid is added to sodium carbonate, two reactions take place.

What are these and how will the titration curves look like?

[RazzMeTazz]

I have a question referring to notation used in chemistry.

Let's say you have a question in which the concentration of a HCl solution is 0.10M and you have 20.00mL of it. The question asks you to find the number of mol of HCl solute.

Would you write:

1.) n(HCl) = c(HCl) × V(HCl)

or

2.) n(HCl) = [HCl] × V(HCl)

So my main question is: Can you use square brackets to indicate the concentration of a solution? Would it be more correct to do this than to say c(HCl) to indicate concentration...?

Thanks in advance!

[keltingmeith]

So my main question is: Can you use square brackets to indicate the concentration of a solution? Would it be more correct to do this than to say c(HCl) to indicate concentration...?

Thanks in advance!

Both are correct and fine to use as long as you're clear about it.

[stockstamp]

This is true, but you can't just assume it - why do you think this is the case?

This is true - but HOW do you know that it's true?

Anyway, I've had a look through your working a bit more thoroughly (sorry for not doing this earlier, I must seem like a dick. >.<), and I think you've just made an arithmetic error. Specifically, (1/2)*0.027603=0.01338015. Otherwise, it looks right to me.

I don't know how else to work out the question - to me, basic logic would suggest that if you want something to be neutral the positives and negatives must cancel - or in this case the acidic and basic components must cancel. If it's wrong to make that assumption, how should I go about solving this question?

Technically it's not true. HSO₄^- is a weak acid with a K_a of around 0.01, so it's not actually fully deprotonated in water. A 1 M sulfuric acid solution has a proton concentration significantly closer to 1 M than it does to 2 M. The only reason why we deprotonated it twice is because it's reacting with a strong base (and because it's a much stronger acid than most other weak acids out there). When reacting with sodium hydroxide, nearly all acids become fully deprotonated (exceptions include the really weak ones out there like NH₄⁺ and ethanol).

I see what you are saying but  K_a values don't come into play until unit 4 so I havent yet learned this. But even so, this brings me back to my former question -  assuming I did take the  K_a value into account - how should I work out the question? That is the only way I can think to do it.

Just in case I'm giving the wrong impression here - I'm not at all arguing with what you are saying or anything like that - I really appreciate the responses but I'm still slightly confused.
Thanks

(I should probably add that this is the only way I can think of to solve all acid/base, titration and back titration questions - so clearly I'm starting to think I've got the wrong approach)

[keltingmeith]

I don't know how else to work out the question - to me, basic logic would suggest that if you want something to be neutral the positives and negatives must cancel - or in this case the acidic and basic components must cancel. If it's wrong to make that assumption, how should I go about solving this question?

Your answer is in the self-ionization of water. However, this equilibrium assumes complete dissociation in water, which only holds for strong acids and bases. So yes, for strong acids and bases, you can assume that 1:1 transfer, but for weaker acids and bases not all of the hydrogens are gathering because some are sticking to their anion, and so you cannot assume 1:1 stoichiometry.