Post by: #1procrastinator on October 28, 2011, 03:48:26 pm
I wrote Cu(s) + AgNO3(aq) -> CuNO3(aq) + Ag+(aq), but the answer is for copper(II) and they have 2Ag(s). Why? Can Ag (just one) be a solid?
Post by: Lasercookie on October 28, 2011, 03:59:00 pm
Write an equation for strip of copper + silver nitrateIf you balance the equation, this is what I got:
I wrote Cu(s) + AgNO3(aq) -> CuNO3(aq) + Ag+(aq), but the answer is for copper(II) and they have 2Ag(s). Why? Can Ag (just one) be a solid?
The copper has a charge of 2+, nitrate ions have a charge of -1. So you need the two nitrates on the right side.
Cu(s) + AgNO3(aq) ---> Cu(NO3)2(aq) + Ag(aq)
This means that you need two nitrates on the left side (and therefore two Ag's on the right side):
Cu(s) + 2AgNO3(aq) ---> Cu(NO3)2(aq) + 2Ag(aq)
So yeah, I'm pretty sure this one here is correct and properly balanced:
Cu(s) + 2AgNO3(aq) ---> Cu(NO3)2(aq) + 2Ag(aq)
Post by: #1procrastinator on October 28, 2011, 04:07:43 pm
Yeah that looks right (compared to the book, which seems to have a lot of errors by the way), except they said silver is a solid (next question asks for mass of it).
Post by: REBORN on October 28, 2011, 04:14:50 pm
Post by: Greatness on October 28, 2011, 04:23:30 pm
Post by: Lasercookie on October 28, 2011, 04:40:16 pm
The question will normally tell you if it's Cu (I), (II), (III) same with Fe, Zn etc. Well i think exam questions do anyway.Yeah I noticed that the textbook most of the time indicates if it's (I), (II), or (III) that they're talking about.
Sorry, I didn't mean to imply that those other forms of Copper doesn't exist :/
Post by: #1procrastinator on October 31, 2011, 01:48:19 pm
No problemo man
I'm still wondering why it's copper(II) and not (I) ...does it have anything to do with stuff from Unit 1? Or is it 'beyond the scope of the course' :p
Post by: Lasercookie on October 31, 2011, 04:56:52 pm
If you assign oxidation numbers to the atoms in the equation - for a reaction to occur, there must have been a change in oxidation number somewhere (is this correct for all situations?).
Note that the oxidation states don't change and therefore a reaction didn't occur? I don't know if this is the correct reasoning. (couldn't think of a better way of writing them, normally I write oxidation numbers directly above - can't really do this when typing)
But, if you consider this:
There is a change in oxidation numbers and therefore a reaction occurs. Silver has an oxisation number of +1 and there's two atoms of silver. While on the right, they have a charge of 0 (since they are in free elemental form or whatever). You can't have electrons being gained without it coming from somewhere else. These come from the copper. The copper atom loses 2 electrons which are received by the silver.
That's the best way I can explain it (in a way that makes sense to me). I am not entirely sure if what I said about the first situation is correct. There's probably a better way of explaining it.
edit: made a mistake, explanation doesn't work - ignore it.
Post by: dinosaur93 on October 31, 2011, 05:03:44 pm
Could anyone help me with these sets of Qs?
Calculate the pH for each of the following solutions.
1. 375.00 mL 0.05 M H2SO4 solution
2. 125.00 mL 0.0005 M HCl solution
3. 350.00 mL 0.01 M NaOH solution
kindly also pls provide self-explanatory answers, tnx heaps :D
Post by: b^3 on October 31, 2011, 05:13:45 pm
If you assign oxidation numbers to the atoms in the equation - for a reaction to occur, there must have been a change in oxidation number somewhere (is this correct for all situations?).Just note for a redox reaction there must be a change in oxidation number.
edit: made a mistake, explanation doesn't work - ignore it.
Post by: Lasercookie on October 31, 2011, 05:17:26 pm
ThanksIf you assign oxidation numbers to the atoms in the equation - for a reaction to occur, there must have been a change in oxidation number somewhere (is this correct for all situations?).Just note for a redox reaction there must be a change in oxidation number.
edit: made a mistake, explanation doesn't work - ignore it.
Would you happen to know the reason behind why Cu(II) is formed and not Cu(I)? I can't find a reason that explains it satisfactorily.
Post by: b^3 on October 31, 2011, 05:19:02 pm
Regarding pH calculation for StoichiometryI'll give you the process to do it, that way you should be able to do it yourself.
Could anyone help me with these sets of Qs?
Calculate the pH for each of the following solutions.
1. 375.00 mL 0.05 M H2SO4 solution
2. 125.00 mL 0.0005 M HCl solution
3. 350.00 mL 0.01 M NaOH solution
kindly also pls provide self-explanatory answers, tnx heaps :D
For pH we need to know the concentration of H+ ions.
1. Find the amount of mol of what you are given (i.e.H2SO4, HCl e.t.c.) using c=n/v
2. From that find out the amount of H+ ions (in mol) using ratios (i.e. 2 mol of H+ for every mol of H2SO4, 1 mol of H+ for every mol of HCl)
2b. If it is a base, you find out the amount of Oh- ions you have using the same method, then (assuming at 25 degrees) use c(OH-)*c(H+)=10^-14 to find the concentration of H+ ions.
3. Find the concentration of H+ ions, i.e. use c=n/v
4. Find the pH using pH=-log(concentation of H+ ions)
Post by: b^3 on October 31, 2011, 05:20:12 pm
I'm not sure but I had a look around before and came up with conflicting explanations. I think we'll need some of the chem masters to help here *looks at Thushan and Mao* (or *looks at Mao and Thushan* if you like it that way)ThanksIf you assign oxidation numbers to the atoms in the equation - for a reaction to occur, there must have been a change in oxidation number somewhere (is this correct for all situations?).Just note for a redox reaction there must be a change in oxidation number.
edit: made a mistake, explanation doesn't work - ignore it.
Would you happen to know the reason behind why Cu(II) is formed and not Cu(I)? I can't find a reason that explains it satisfactorily.
Post by: dinosaur93 on October 31, 2011, 05:21:19 pm
Post by: b^3 on October 31, 2011, 05:23:11 pm
dude, Ive tried doing it already, the thing that doesn't make sense to me though is the fact that I'm ending up with a negative pH and doesn't know how to proceed with that value to find an answer.Slightly negative ph's are possible but I'll give it a go and post back in a min.
Post by: Lasercookie on October 31, 2011, 05:24:25 pm
dude, Ive tried doing it already, the thing that doesn't make sense to me though is the fact that I'm ending up with a negative pH and doesn't know how to proceed with that value to find an answer.pH=-log(concentation of H+ ions)
Are you forgetting that the equation has a negative at the front? You need to put that in to get a positive answer.
(edit: unless this is one of those negative ph situations b^3 is talking about)
Post by: b^3 on October 31, 2011, 05:27:31 pm
Answers you should be getting are
1. pH=1
2. pH=3.3
3. pH=12
If you got the -ve of those then you may have forgotten the -ve.
Post by: #1procrastinator on November 04, 2011, 11:42:52 am
I know oxygen is -2 but Na is +1 and Cl is -1. Which one do I change and why? i.e. Na is +3 and Cl is -1 or both Na and Cl are +1
Post by: Lasercookie on November 04, 2011, 08:56:45 pm
Find the oxidation numbers of each of the elements in this compound NaClOThe two ions there are Sodium (Na) and Chlorite (ClO).
I know oxygen is -2 but Na is +1 and Cl is -1. Which one do I change and why? i.e. Na is +3 and Cl is -1 or both Na and Cl are +1
Chlorite has a charge of 1-
So to show you the working out:
x - 2 = -1
x = -1 + 2
x = +1
So Chlorine must have a charge of +1.
edit: Chlorite is an incorrect name for that ion. It's actually called hypochlorite.
http://en.wikipedia.org/wiki/Chlorite
As you can see though, they all have the overall charge of 1- though.
Post by: dinosaur93 on November 09, 2011, 04:34:05 pm
In the above reaction, the H2PO4- ion acts as:
a. an acid
b. an amphiprotic substance
c. a reductant
d. an oxidant
Why is the answer b alone?
Post by: Lasercookie on November 09, 2011, 04:44:32 pm
2H2PO4-Well it's amphiprotic, so it can't be A.HPO42- + H3PO4-
In the above reaction, the H2PO4- ion acts as:
a. an acid
b. an amphiprotic substance
c. a reductant
d. an oxidant
Why is the answer b alone?
If you have a reductant, you need something else to be an oxidant. Redox reactions must have both happening simultaneously.
Remember that with acid/base, you have protons being donated/accepted - but electrons are shared. With redox reactions, electrons are being transferred.
Post by: dinosaur93 on November 09, 2011, 04:51:24 pm
Calculate the pH of the solution.
m ( H2SO4 ) = 12.5g
N ( H2SO4 ) = 800 mL
n ( H2SO4 ) =
n ( H2SO4 ) =
n ( H2SO4 ) = 0.127395 mol
c ( H2SO4 ) =
c ( H2SO4 ) =
From this stage on, the solution pathway says:
pH = -log [H3O+ = -log [0.159] = 0.799
but isnt we still need to multiply the concentration by 2 (before solving for pH) because there is 2 H+ ions in the given equation? ???
Post by: dinosaur93 on November 09, 2011, 04:53:59 pm
right! I was thinking of reductant, but in this case since the solution is not mixed or reacted with another solution, therefore it cannot satisfy the redox reaction, is that right? tnx bro!2H2PO4-Well it's amphiprotic, so it can't be A.
In the above reaction, the H2PO4- ion acts as:
a. an acid
b. an amphiprotic substance
c. a reductant
d. an oxidant
Why is the answer b alone?
If you have a reductant, you need something else to be an oxidant. Redox reactions must have both happening simultaneously.
Remember that with acid/base, you have protons being donated/accepted - but electrons are shared. With redox reactions, electrons are being transferred.
Post by: #1procrastinator on November 10, 2011, 11:52:25 am
Find the oxidation numbers of each of the elements in this compound NaClOThe two ions there are Sodium (Na) and Chlorite (ClO).
I know oxygen is -2 but Na is +1 and Cl is -1. Which one do I change and why? i.e. Na is +3 and Cl is -1 or both Na and Cl are +1
Chlorite has a charge of 1-
So to show you the working out:
x - 2 = -1
x = -1 + 2
x = +1
So Chlorine must have a charge of +1.
edit: Chlorite is an incorrect name for that ion. It's actually called hypochlorite.
http://en.wikipedia.org/wiki/Chlorite
As you can see though, they all have the overall charge of 1- though.
Thanks, teacher treated each ion separately
Post by: dinosaur93 on November 10, 2011, 07:20:59 pm
Is it PO42- or H2PO4?
Post by: REBORN on November 10, 2011, 07:22:06 pm
Post by: dinosaur93 on November 10, 2011, 07:24:45 pm
because the solution pathway that I had says H2PO4!
tnx heaps bro!