# Molarity notation

1. Apr 5, 2005

### preet

What does the following notation (to the power of -1) mean? TiA

ex. $$[Na] = 0.050 mol L ^{-1}$$

2. Apr 5, 2005

### preet

I have another question (did not want to create a new thread):

"What is the ratio x:y when the equation below is properly balanced?"

$$xSn^{2+}(aq) + y Ag^{+}(aq) -> n Sn^{4+}(aq) + m Ag^{+}(s)$$

I've never seen a question like this before... an explanation or a link to a site or something would be greatly appreciated.

3. Apr 5, 2005

### pack_rat2

"L^-1" means "per liter." 0.05 mol/L is 0.05 M.

4. Apr 5, 2005

### dextercioby

For the first post,it's simply the unit 'liter' (which should be shortened 'l',not 'L' (that stands for length)) raised to the power "-1".

For the second,i'm sure u miss the negative ions...Silver ion is a spectator in a redox ionic reaction.I don't see a connection between "x" & "y".And next time use $\rightarrow$ (code \rightarrow).

Daniel.

5. Apr 5, 2005

### pack_rat2

Simply balance the equation and give the ratio of x to y. This is a re-dox reaction. They're usually solved using the method of "half-reactions."

6. Apr 5, 2005

### pack_rat2

Ooooops! I thought that was "Ag," not "Ag+"....and there is something wrong, here. Sn(+4) + 2Ag -> Sn(+2) + 2Ag(+) might be the reaction, but not what you have.

7. Apr 5, 2005

### dextercioby

That's what i said above and it seemed weird to me,too that "Ag" doesn't undergo either reduction or oxdation.

Daniel.

8. Apr 5, 2005

9. Apr 5, 2005

### dextercioby

$$Sn^{2+}_{(aq)} + Ag^{+}_{(aq)}\rightarrow Sn^{4+}_{(aq)}+Ag\downarrow$$

Now you can do the redox properly...

Daniel.

10. Apr 6, 2005

### Staff: Mentor

11. Apr 6, 2005

### dextercioby

Nope,it's incorrect.It's the same way they teach "k" instead of "K" for Kg...It's outrageous.

Daniel.

12. Apr 6, 2005

### Staff: Mentor

Have you read comment on the NIST page? L was adopted in 1979 and is internationally accepted. So it is correct.

IUPAC lists both forms just like NIST does:

http://www.iupac.org/reports/1993/homann/units51.html

I was taught l 30 years ago and I am not advocating L - but it seems L is now accepted by all major institutions and I must agree with the fact that L is much less prone to be mistaken with 1 then l is. It doesn't mean I like it :)

Chemical calculators for labs and education
BATE - pH calculations, titration curves

13. Apr 6, 2005

### pack_rat2

I usually use "L" for liter, and "ml" for milliliter. When on a computer or on the Net where certain specific fonts are employed, I *HATE* to use "l" because it looks too much like "I".

14. Apr 6, 2005

### dextercioby

I explained the reasoning with "l" vs."L".Capitals are used for physical quantities and multiples.Liter is not a part of the units which use capitals...I'm sorry for the French,but they're wrong.

Daniel.

15. Apr 7, 2005

### Staff: Mentor

m stands for mili and M for Mega - both are SI multiples.

s stands for second, K for Kelvin - both are base SI units.

So either I don't understand what you have written or you are not right

What I am aiming at is that there are no 'hard' rules.

And, while we can criticise units abbreviations defined by international organizations like IUPAC or CGPM we have no choice but to accept them (and to fight for changes if we think it is important)

Chemical calculators for labs and education
BATE - pH calculations, titration curves

16. Apr 7, 2005

### dextercioby

Sorry,i wasn't really precise.It happens from time time,i'm human,though.I may be wrog,occasionally.

Daniel.

17. Apr 13, 2005

### preet

Im sorry to bring this post up but I wanted to be sure that I was doing the redox equation right...

$$Sn^{2+}_{(aq)} + Ag^{+}_{(aq)}\rightarrow Sn^{4+}_{(aq)}+Ag$$

was the equation (sorry for my typo earlier)

And I needed to find the coeffecient on the left side of the equation (in front of tin and silver)... so..

*tin has lost two electrons
$$Sn^{2+}\rightarrow Sn^{4+}+ 2e^{-}$$

*silver has gained one electron (one atom of silver has gained one electron)
$$Ag^{+} + e^{-}\rightarrow Ag$$

*2 electrons were lost by tin so to balance that I multiply Ag by 2...

and the coeffecients are 1 in front of Sn and 2 in front of Ag? Is this right? TiA!

18. Apr 13, 2005

### dextercioby

Yes,it is correct...

Daniel.