Gen Chem (Dat Destroyer 2016) #172

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

dentisttobe19

Full Member
7+ Year Member
Joined
Jun 2, 2016
Messages
29
Reaction score
2
I'm confused about this question.. It says If 20 mL of .012M soln. of Ca(OH)2 is added to a 48 mL of HBR, what is the concentration of the HBr?

The answers say not to use M1V1=M2V2. Instead use Normality. SO NaVa=NbVb

Na[48]=.024(20) This is where I'm confused about where the .024 came from because when you find the moles ... you get .012M(20mL)=.24mol and then in the problem they use .024... where did that come from?

Thanks!

Members don't see this ad.
 
I'm confused about this question.. It says If 20 mL of .012M soln. of Ca(OH)2 is added to a 48 mL of HBR, what is the concentration of the HBr?

The answers say not to use M1V1=M2V2. Instead use Normality. SO NaVa=NbVb

Na[48]=.024(20) This is where I'm confused about where the .024 came from because when you find the moles ... you get .012M(20mL)=.24mol and then in the problem they use .024... where did that come from?

Thanks!
At first, you should notice that this is actually a titration between a strong acid and a strong base. M1V1=M2V2 is the formal you use to solve this question. The reason why it is 0.024 and not 0.012 is that there are 2 moles of OH- in each mole of Ca(OH)2 so you have to multiply the concentration of Ca(OH)2 by 2. Normality is just nMV (n is the # of moles of either OH- or H+)
 
I'm confused about this question.. It says If 20 mL of .012M soln. of Ca(OH)2 is added to a 48 mL of HBR, what is the concentration of the HBr?

The answers say not to use M1V1=M2V2. Instead use Normality. SO NaVa=NbVb

Na[48]=.024(20) This is where I'm confused about where the .024 came from because when you find the moles ... you get .012M(20mL)=.24mol and then in the problem they use .024... where did that come from?

Thanks!
Ca(OH)2 is dibasic. This means it has 2 molecules of OH. HBr only donates one proton. That is why we use Normality in this question, since the ratio of H+ to OH- is not one to one.
So, if you take 0.012 M *2 =0.024.

Hope this helps.
 
All elements in group 2 hydroxides donates two OHs, so don't get tricked when they start giving you the names of the elements like"barium hydroxide" and nothing else.
 
Members don't see this ad :)
M1V1=M2V2 is only used for dilutions. N1V1=N2V2 is used when titrating to the equivalence point.
I think this is a more thorough way of solving this problem:

Since we're doing a titration, we'll use N1V1=N2V2. So convert the concentrations into normality:
20mL 0.012M Ca(OH)2 dissociates to 2 mol OH- so the math would be
(0.012M Ca(OH)2)(2eq OH-/mol Ca(OH)2) = 0.024N or 0.024eq OH-/L soln

From there just plug the values into the formula making sure to convert the given volumes to L. This should give you 0.010N (0.010eq H+/L soln).
Since HBr is a monoprotic strong acid, its normality will be the same as its molarity. The math would be
(0.010eq H+/L soln)(mol HBr/1eq H+) = 0.010M HBr
 
Ca(OH)2 is dibasic. This means it has 2 molecules of OH. HBr only donates one proton. That is why we use Normality in this question, since the ratio of H+ to OH- is not one to one.
So, if you take 0.012 M *2 =0.024.

Hope this helps.

I understand all this after reading the explanations, but what part of the question was focused on to infer that this was a titration problem? I understand that it's a strong base and strong acid addition, but couldn't the HBr have just been any concentration theoretically and not just enough to titrate?
 
Top