N vs M in GC.

[babowc]

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Working the destroyer after watching Chad's, I came across this problem:

2011 GC destroyer Q116: If 20ml of 0.012M solution of Ca(OH)2 is added to 48ml of HBr, what's the concentration of the HBr?

I casually used MaVa=MbVb.

I checked the solution and it said I can't use it, because differences in OH and H.

My question is... how often does N/M usage come up on the real DAT?
I can't remember seeing such question in the real thing, nor any of the tests I've taken so thus far.

Thanks

 

[Dotoday]

Normality comes on occasionaly. You can never know exactly what they will ask. I think it is better be ready for anything.

Sent from my SPH-L710 using Tapatalk

 

[babowc]

I see..
So, if it was Ca(OH)3 instead of 2, its concentration is "tripled" for N?

 

[xatlasb]

I would just always use normality, you can't ever go wrong. 🙂

 

[patricklin27]

How do you know when to use normality and when to use molarity?

 

[game33]

How I remember it is.....if they ask you about the dilution than you use M1V1=M2V2. If they ask you about concentration or titrated, or neutralization than you use N1V1=N2V2 and remember that H's matter so N1V1=N2V2. Let me know and I'll give you an explain but there is 2 or 3 good ones in Destroyer.

 

[patricklin27]

How I remember it is.....if they ask you about the dilution than you use M1V1=M2V2. If they ask you about concentration or titrated, or neutralization than you use N1V1=N2V2 and remember that H's matter so N1V1=N2V2. Let me know and I'll give you an explain but there is 2 or 3 good ones in Destroyer.

so basically in these questions when you use the N1V1=N2V2 equation.. if the question doesn't talk about dilution, then we can assume to multiple the concentration and the number of moles of either H+ or OH- for N1 (or N2).

 

[game33]

so for this problem you see that it's asking you about concentration so you automatically should think N1V1=N2V2. First you see that Ca(OH)2 actually has 2 OH hence 2 H's and HBr has only one H. Now you see that (20)(.012)=(N2)(48). It becomes (20)(.024)=(N2)(48) because you multiple .012 X 2 because Ca(OH) 2 has 2 OH hence again 2 H's. Once you get (20)(.024)=(N2)(48) you just do simply division and you get your answer of .01N HBr=.01M HBr. Hope this helps.

 

[babowc]

That helped, but what about this question...?

Q181 Destroyer 2011.
30ml of .20M barium hydroxide Ba(OH)2 is required to neutralize 25ml of citric acid H3C6H5O7. What is the molarity of the citric acid solution?

I did the math, and came up with .550M, but the answer key states that because there are "3 ionizeable protons in citric acid, the answer needs to be divided by 3, thus the answer is .167M

When do you know to divide the answer? when there are free hydrogens?