Calculate the molarity of Br2 in the stock solution.

[mousesgr]

The stock solution of bromine is prepared by dissolving 232.5 g of Br2 and 187.5 g of KBr in water and diluting to 750 ml. Calculate the molarity of Br2 in the stock solution.

how to do this qs?

 

[dextercioby]

What's the definition of molarity...?Theory must be known b4 attempting to solve exercises...

Daniel.

 

[mousesgr]

What's the definition of molarity...?Theory must be known b4 attempting to solve exercises...

Daniel.

no. of mole / volume?

 

[dextercioby]

PER LITER OF SOLLUTION... Now compute it...

Daniel.

 

[Edgardo]

From http://www.webelements.com/webelements/elements/text/Br/key.html
I took the mass of Br. Since you have $\mbox{Br}_{2}$,
one 1mol of $\mbox{Br}_{2}$ weighs 2* 79.904g

Divide 232.5 g by (2* 79.904g) , which equals 1.45487.
So you got 1.45mol of $\mbox{Br}_{2}$.

Now you need the molarity which is the concentration:

$c= \frac{1.45mol}{0.75l} = 1.93 \frac{mol}{l}$

I hope this is right.

I don't know what role the KBr plays, I guess that it helps
dilluting the $Br_{2}$. Note that I didn't take into account
the KBr.

 

[dextercioby]

Even if this is not posted in the HM section,still solving the problem instead of the OP is not reccomendable.So take this as an advice for future "urges"...

Daniel.

 

[Edgardo]

I think there still may be a mistake in my solution.
The fact that they give the information "187.5 g KBr" makes me a little
bit doubtful.

If you have to take the KBr into account, then I hope you know
how to solve it mousesgr.

 

[dextercioby]

That "Br2" (or $Br_{2}$) is surely misleading.It may still refer only to bromine MOLECULES...But if u read it simply as "bromine" (as in "molarity of bromine"),then,of course,u have to include the bromine from the bromide.

Danie.

 

[Gokul43201]

The stock solution of bromine is prepared by dissolving 232.5 g of Br2 and 187.5 g of KBr in water and diluting to 750 ml. Calculate the molarity of Br2 in the stock solution.

how to do this qs?

I've been trying to not answer this because it's not clear what the level of the poster is. At the simplest level, you just do the calculation performed by edgardo.

At a higher level of sophistication, you need to consider what happens when you dissolve Bromine in water...about the same that happens with chlorine :

[tex]Br_2 + H_2O \leftrightharpoons HBr + HOBr [/itex] But it's important to note that the products will be present in aqueous solution, in their ionic forms, $H^+,~Br^-,~Br^+,~OH^-$. The amounts of molecular and ionic bromine can be calculated easily from the equilibrium constant for this reaction. The effect of adding KBr is that it produces $Br^-$ ions in solution, shifting the equilibrium leftwards, resulting in more molecular bromine. Assuming KBr is highly ionic (ie : the dissociation constant is close to 1), one can then calculate $[Br_2]_{aq}$.

 

[Borek]

But it's important to note that the products will be present in aqueous solution, in their ionic forms, $H^+,~Br^-,~Br^+,~OH^-$.

$~Br^+$? Weak as it is (pKa = 8.7) $~HOBr$ is an acid. Look for $~OBr^-$.


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[Gokul43201]

$~Br^+$? Weak as it is (pKa = 8.7) $~HOBr$ is an acid. Look for $~OBr^-$.


Borek

Yes, it is; isn't it (duh) ? My bad !

 

[Borek]

Yes, it is; isn't it (duh) ? My bad !

Nobodys perfect... It is www.chembuddy.com, .pl was left from other project


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