Bazza's 3/4 chemistry questions

[WhoTookMyUsername]

yeah i'm pretty sure that's exactly what i'm asking, thanks !

[WhoTookMyUsername]

Is there any particular reason why a standard solution is usually placed in the burette?

[Hamdog17]

Think about it, the standard solution generally used is of LOWER concentration then the analyte hence by having precise control over the volume of standard solution reacted we can more accurately determine the concentration rather than if it were performed the other way. Also a convenience thing, if your standardising unknown bases then it would be helpful if your burette only ever had one solution in it during the experiment- the standard acid- it would also reduced contamination due to the bases mixing etc.

[Shenz0r]

Wouldn't the standard solution be the limiting reagent in the reaction? Hence we can more accurately measure how much of it was involved in the reaction if we measured it with the burette?

[WhoTookMyUsername]

Think about it, the standard solution generally used is of LOWER concentration then the analyte hence by having precise control over the volume of standard solution reacted we can more accurately determine the concentration rather than if it were performed the other way. Also a convenience thing, if your standardising unknown bases then it would be helpful if your burette only ever had one solution in it during the experiment- the standard acid- it would also reduced contamination due to the bases mixing etc.

ah yes that makes perfect sense, thanks

another question

for a reaction involving a potentially polyprotic acid such as sulfuric acid, how do we know if it will ionise completely?
(e.g. whether it will remain HSO4 or go to SO4)
my question is specifically related to the NaOH + H2SO4 reaction, which i thought woudl form NaHSO4 + H2O but doesn't according to the text ?

[Aurelian]

Think about it, the standard solution generally used is of LOWER concentration then the analyte hence by having precise control over the volume of standard solution reacted we can more accurately determine the concentration rather than if it were performed the other way. Also a convenience thing, if your standardising unknown bases then it would be helpful if your burette only ever had one solution in it during the experiment- the standard acid- it would also reduced contamination due to the bases mixing etc.

ah yes that makes perfect sense, thanks

another question

for a reaction involving a potentially polyprotic acid such as sulfuric acid, how do we know if it will ionise completely?
(e.g. whether it will remain HSO4 or go to SO4)
my question is specifically related to the NaOH + H2SO4 reaction, which i thought woudl form NaHSO4 + H2O but doesn't according to the text ?

In the presence of a strongly basic solution such as NaOH (aq), a diprotic acid will generally be completely ionized (provided NaOH is in excess...).

In other questions concerning polyprotic acids (e.g. U3 pH determination), you will generally be given information regarding to what extent ionization occurs.

[WhoTookMyUsername]

thanks aurelian

I have trouble deciphering which solution is being referred to.
When it says something like "excess sodium hydroxide in the resulting solution was titrated with 0.1132 M HCl)


does the "titrated with" imply that the HCL is the titre?

[Aurelian]

thanks aurelian

I have trouble deciphering which solution is being referred to.
When it says something like "excess sodium hydroxide in the resulting solution was titrated with 0.1132 M HCl)


does the "titrated with" imply that the HCL is the titre?

Probably... but not necessarily. It's important to remember that when doing titration, it actually doesn't matter what's in the burette and what's in the conical flask. Here since you know the concentration of the HCl (aq) but not the NaOH (aq), evidently you're trying to find the concentration of the latter. However, you could just as easily be releasing HCl (aq) from the burette into a known volume of the NaOH (aq) as you could be releasing NaOH (aq) into a known volume of the HCl (aq). It doesn't generally really matter; "titrated with" really just means x and y were involved in a titration experiment with each other.

I'm not sure if that answers your question though haha

[WhoTookMyUsername]

What exactly is a free element? Google didnt bring a conclusive answer

[WhoTookMyUsername]

yeah that sounds right, thanks

um
i have no idea what the text is doing

7a) write ionic half equations for the:
reduction of MnO₂ to Mn²⁺.

The answer is
MnO₂(s) + 4H⁺(aq) + 2e^- -> Mn²⁺(aq) + 2h_2o (l)

now i have 2 problems with this answer; first, i don't think it's in half equation form, secondly, where do the random hydrogen atoms come from? it hasn't stated it's an acidic solution ?

thanks (if this is wrong, correct answer would be appreciated)

[thushan]

its a half eqn - there are free electrons in the equation
yeah we generally assume acidic media unless otherwise stated.

[pi]

its a half eqn - there are free electrons in the equation
yeah we generally assume acidic media unless otherwise stated.

That's correct.

What made you think it wasn't in half-equation form? If e- are present, the equation is definitely a half-equation

[WhoTookMyUsername]

it's just whenver i did half equations it was always like

Na -> Na+ + e or something

[pi]

it's just whenver i did half equations it was always like

Na -> Na+ + e or something

Well, that's a simple half-equation. The more complex ones have H+ or OH- in them (and of course H2Os) You'll see a few in your data book and more when your titrating with substances that have dichromate ions, etc.

[Aurelian]

yeah that sounds right, thanks

um
i have no idea what the text is doing

7a) write ionic half equations for the:
reduction of MnO₂ to Mn²⁺.

The answer is
MnO₂(s) + 4H⁺(aq) + 2e^- -> Mn²⁺(aq) + 2h_2o (l)

now i have 2 problems with this answer; first, i don't think it's in half equation form, secondly, where do the random hydrogen atoms come from? it hasn't stated it's an acidic solution ?

thanks (if this is wrong, correct answer would be appreciated)

There is a specific method for balancing 'complex' half-equations such as these. Do you have Heinmann Chemistry 2 Enhanced 4th Edition? If so, flip to page 54 and read "5.3 Writing half equations" for an explanation