Have I been thinking of Normality incorrectly?

[bleach121]

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This question is from BR Gen Chem Ch 5 P8 # 50:
If Flask 1 requires exactly 20.0 mL NaOH solution to be neutralized, what must be the concentration of the NaOH solution? (Given that 25.0 mL of 0.20 N H2SO4 are already present in the flask.)

1. 0.125 MNaOH(aq)
   
   
2. 0.200 M NaOH(aq)
   
   
3. 0.250MNaOH(aq)
   
   
4. 0.500MNaOH(aq)

The answer is 0.25 M. I would appreciate it if someone could explain to me what I did wrong, and if it is okay to assume that Normality = (Molarity) x (# of equivalents) for a problem like this.

Here's what I did: to get molarity of H2SO4 from the given Normality, I did 0.2/2 = 0.1 M H2SO4. Then I used M1V1=M2V2.
0.1M H2SO4 (.025L H2SO4) = .02L NaOH (X)
X = .125 M NaOH

Thanks

 

[Czarcasm]

This question is from BR Gen Chem Ch 5 P8 # 50:
If Flask 1 requires exactly 20.0 mL NaOH solution to be neutralized, what must be the concentration of the NaOH solution? (Given that 25.0 mL of 0.20 N H2SO4 are already present in the flask.)

1. 0.125 MNaOH(aq)
   
   
2. 0.200 M NaOH(aq)
   
   
3. 0.250MNaOH(aq)
   
   
4. 0.500MNaOH(aq)

The answer is 0.25 M. I would appreciate it if someone could explain to me what I did wrong, and if it is okay to assume that Normality = (Molarity) x (# of equivalents) for a problem like this.

Here's what I did: to get molarity of H2SO4 from the given Normality, I did 0.2/2 = 0.1 M H2SO4. Then I used M1V1=M2V2.
0.1M H2SO4 (.025L H2SO4) = .02L NaOH (X)
X = .125 M NaOH

Thanks

You have the right idea, but you made a mistake though. If you choose to divide the normality out to find molarity, then use the equation: iM1V1 = iM2V2, because in each case, we have a different number of equivalents (a mistake frequently overlooked by the way). For every 1 mole OH- (i=1), we have 2 moles H+ (i=2). By dividing to find molarity, you essentially found the number of moles to neutralize half the solution. You would need twice as much NaOH to compensate (twice the concentration) and you can achieve this by either halving the volume of solvent or doubling the the number of moles.

Another approach to make things even is to find the normality of NaOH rather than dividing the normality of H2SO4 (to find its molarity). This is what I generally use: N1V1 = N2V2. Then I divide 'N' by i afterwards to find molarity (and multiply this by total volume in L to find moles - if asked).

 

[orangetea]

I used to make the same mistake! But what I realized is that normality actually accounts for the equivalents. Just remember normality's units is Eq/L. 😀

p.s. Eq= equivalents

 

[bleach121]

Great, thanks guys!