Chemistry Questions (3 Viewers)

[madharris]

Re: Determining acidity/alkalinity of a salt

K2SO4 is actually a basic salt.

Even though H2SO4 is a strong acid, only the first ionisation is strong.

Oh yeah I forgot about that!, thanks

And madharris, can I just ask why the equilibrium will shift left if KOH is a strong base? Is it because that it has a great tendency to attract a proton, hence it does this and forms the reactants, causing the equib to shift left?

Yeah I'm pretty sure that's why

 

[Aysce]

Acid-base Titration claification

Alright in Conquering Chemistry it reads:

Fill a burette (Figure 8.5 on p. 215 CCPC) with a solution of known concentration (for example sulfuric acid) and adjust the solution level in the burette to the zero mark†; the solution in the burette is called the titrant.

Alright so in this case, the solution of known concentration in the burette is the standard solution since we know it's exact concentration. Although, a standard solution is gained from a primary standard and within a primary standard, we are unable to use a substance such as sulfuric acid since when it is diluted, the concentration changes over time and reduces the accuracy of our results ie. We need a substance that is sufficiently stable and pure.

So can anyone clarify why the textbook would use a solution of sulfuric acid as the standard solution? Sorry if it is a little confusing.

EDIT:

An additional question that I need confirmation for.

According to Conquering Chemistry, a primary standard is the ACTUAL substance (with sufficient purity and stability) that is dissolved in a certain volume of water to form the standard solution.

However, according to Kamal's notes, it is stated that a primary standard is the solution made from scratch by dissolving the substance into a certain volume of water.

So I'm a little caught up between those two as well.

Thanks so much!

 

[nightweaver066]

Re: Acid-base Titration claification

Using sulfuric acid is fine in LOW concentrations. At higher concentrations, it becomes more hygroscopic, i.e., it absorbs water and thus changes in concentration.

The primary standard is the actual substance, not the solution.

Now get off and go partyyyy

 

[Aysce]

Re: Acid-base Titration claification

Using sulfuric acid is fine in LOW concentrations. At higher concentrations, it becomes hygroscopic, i.e., it absorbs water and thus changes in concentration.

The primary standard is the actual substance, not the solution.

Now get off and go partyyyy

Alright, thank you

And yeah brah dw

 

[Aysce]

Buffer solution question

Just need some confirmation here since there are no answers.

Q. Would 1 L of aqueous solution containing each of the following pairs of substances be a buffer solution or not? Explain why (or why not). Use an equation where appropriate.

d. 0.25 mol methanoic (formic) acid and 0.25 mol sodium methanoate

So writing down the equation:

CHCOOH + NaOH ----> H2O + Na^+ + CHCOO^-

So we know that the formic acid is generally a weak acid and when it reacts with sodium hydroxide, it produces sodium methanoate (Ionic form above). So would the answer be yes? Justification: Since formic acid is a weak acid and the conjugate base of this acid: CHCOO^- are both present, this aqueous solution will be a buffer solution.

Yes or no?

Thanks!

 

[golgo13]

Re: Buffer solution question

For the first part it would be true if you are saying that the pair is a weak (acid/base) and it's corresponding conjugate. Use human blood buffer (H2CO3/HCO3-). If however strong (acid/base) and is used say (HCL/Cl-) it is not a buffer. Reasons everythings about reasons. Recall the degree of ionisation so strong acid and strong base have a higher degree of ionisation, meaning if there was change they'll jump straight to one side no question. Whereas with weak acid/base their degree of ionisation is lower, meaning they are sorta centre leaning either left or right but will take some time and is quite happy fence sitting until it's really persuaded by peer preasure (large amounts of acid/base depending)

SECOND PART
I'm not sure, but i don't think it's a buffer system, just a hunch that the Na is some role in the reason i think other members will be able to answer this with much more solidity than this

 

[Aysce]

Re: Buffer solution question

For the first part it would be true if you are saying that the pair is a weak (acid/base) and it's corresponding conjugate. Use human blood buffer (H2CO3/HCO3-). If however strong (acid/base) and is used say (HCL/Cl-) it is not a buffer. Reasons everythings about reasons. Recall the degree of ionisation so strong acid and strong base have a higher degree of ionisation, meaning if there was change they'll jump straight to one side no question. Whereas with weak acid/base their degree of ionisation is lower, meaning they are sorta centre leaning either left or right but will take some time and is quite happy fence sitting until it's really persuaded by peer preasure (large amounts of acid/base depending)

SECOND PART
I'm not sure, but i don't think it's a buffer system, just a hunch that the Na is some role in the reason i think other members will be able to answer this with much more solidity than this

Yeah I'm a little unsure if Na has an actual role in the buffer system, or if the CHCOO- ion is just considered and the Na+ from CHCOONa is ignored.

 

[Aysce]

Re: Buffer solution question

Bump.

 

[Aysce]

Re: Buffer solution question

.

 

[Trebla]

Re: Buffer solution question

Yeah I'm a little unsure if Na has an actual role in the buffer system, or if the CHCOO- ion is just considered and the Na+ from CHCOONa is ignored.

Na+ is just a spectator ion. It plays no part in the reaction since it remains an ion on either side.

 

[Aysce]

Re: Buffer solution question

Na+ is just a spectator ion. It plays no part in the reaction since it remains an ion on either side.

...Well this is awk..

Thanks!

 

[someth1ng]

Re: Buffer solution question

You can consider that the Na+ has a role but it is not in the actual reaction as Trebla has stated as it is a spectator ion. The Na+ is only there to neutralise the negative charge of the HCOO- ions.

Remember the definition of a buffer - a solution containing roughly equal amounts of a weak acid and its conjugate base.

Adding NaOH CAN make it a buffer but just because you added NaOH does not mean that the final solution will be a buffer.

 

[Aysce]

Re: Buffer solution question

You can consider that the Na+ has a role but it is not in the actual reaction as Trebla has stated as it is a spectator ion. The Na+ is only there to neutralise the negative charge of the HCOO- ions.

Remember the definition of a buffer - a solution containing roughly equal amounts of a weak acid and its conjugate base.

Adding NaOH CAN make it a buffer but just because you added NaOH does not mean that the final solution will be a buffer.

Alright, I'll keep that in mind. Thanks

 

[Aysce]

Equilibrium expression and constant question



Alright, I got this question right but the intuition behind my method isn't clear to me.

Okay so here's the working out:

<a href="http://www.codecogs.com/eqnedit.php?latex=K = \frac{[H2O][CO]}{[H2][CO2]}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?K = \frac{[H2O][CO]}{[H2][CO2]}" title="K = \frac{[H2O][CO]}{[H2][CO2]}" /></a>

1 mol of H2 and 1 mol of CO2 was placed inside.

Concentration of CO2 = 0.44 mol/L

Using the 1:1 ratio of the reactants, [H2] = 0.44 mol/L

Okay so here's the part where I'm a little unsure:

So as seen, 1 mol/L of H2 and CO2 has been decomposed to 0.44 mol/L for each respectively to form the products. Though from here I mindlessly associated 0.56 mol/L to the products and finished off the question.

I don't understand why the products would have a molarity of 0.56 mol/L?

Thank you!

 

[DamTameNaken]

Re: Equilibrium expression and constant question

You have your equilibrium reaction: H2 + CO2 <--> H2O + CO

You only start the one mole of H2 and the one mole of CO2, so it takes a while for the system to reach equilibrium.

Since we only start with one mole of H2, then Moles of H2 + moles of H20 = 1. Similarly Moles of CO2 + Moles of CO = 1.

Once the system has reached equilibrium you work out that you have 0.44 mol of CO2 and therefore 0.56 mol of CO.

Since H2 and CO2 are both reactants, if we have 0.44 mol of CO2 we also have 0.44 mol of H2. Therefore we have 0.44 mol of H2 and 0.56 mole of H20.

Now we can work out the answer and I've sort of addressed your confusion, but i'll explain it again in different words, just in case.

We start out with 1 mol of H2 and 1 mol of CO2 in a space of 1L. This means we have 1 mol/L of H2 and CO2. We don't have any H2O or CO2 until the equilibrium starts. The equilibrium settles down and we're left with 0.44 mol of CO2 meaning 0.56 mol of CO2 has been transformed somewhere i.e. into CO. Likewise for H2 and H20.

 

[Aysce]

Re: Equilibrium expression and constant question

You have your equilibrium reaction: H2 + CO2 <--> H2O + CO

You only start the one mole of H2 and the one mole of CO2, so it takes a while for the system to reach equilibrium.

Since we only start with one mole of H2, then Moles of H2 + moles of H20 = 1. Similarly Moles of CO2 + Moles of CO = 1.

Once the system has reached equilibrium you work out that you have 0.44 mol of CO2 and therefore 0.56 mol of CO.

Since H2 and CO2 are both reactants, if we have 0.44 mol of CO2 we also have 0.44 mol of H2. Therefore we have 0.44 mol of H2 and 0.56 mole of H20.

Now we can work out the answer and I've sort of addressed your confusion, but i'll explain it again in different words, just in case.

We start out with 1 mol of H2 and 1 mol of CO2 in a space of 1L. This means we have 1 mol/L of H2 and CO2. We don't have any H2O or CO2 until the equilibrium starts. The equilibrium settles down and we're left with 0.44 mol of CO2 meaning 0.56 mol of CO2 has been transformed somewhere i.e. into CO. Likewise for H2 and H20.

Ah thank you very much DamTameNaken!

May I ask why you're already up to Industrial Chemistry?

 

[Aysce]

Chemistry Calculation questions

Hello everyone!

Can anyone recommend to me/send me resources that mostly or purely consist of calculation questions?

Thanks a lot.

 

[DamTameNaken]

Re: Equilibrium expression and constant question

I was under the impression that the question came from the 'acidic environment' module and I just used the theory from 'AE' to answer the question. I'm not yet up to Inductrial Chem to answer your question.

 

[Aysce]

Re: Equilibrium expression and constant question

I was under the impression that the question came from the 'acidic environment' module and I just used the theory from 'AE' to answer the question. I'm not yet up to Inductrial Chem to answer your question.

Yeah I kinda realised my question wasn't anything too specific or restricted to a module ^____^'

 

[Aysce]

Re: Chemistry Calculation questions

Bump.