DAT Achiever Test 1 GC #41 Help!

[dkidemo]

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How many milliliters of 3.0M H2SO4 are required to neutralize 30.0 ml of 4.0 M NaOH?

I always learned that if you have a strong base with a strong acid the neutralization is at pH=7, and to get that you find the equivalence point. In the achiever answer they say that it is obvious that only 0.5 mole of H2SO4 is required to Neutralize every mole of NaOH. I am not seeing how this is obvious and I want to know how I should be processing this question come test time. The answer is 20 ml and I put 40ml.

Thanks

 

[obeserabbit]

I think it's because H2SO4 is a diprotic acid

edit for clarity:
If you use the formula N1V1 = N2V2, you get...
(2)(3.0M)(V) = (4.0M)(30.0ml)
V = 20ml

 

[J777]

Ya, that's right. Since it is a diprotic acid, there are effectively two H+ for each OH-. So only 1/2 = 0.5 mole of H2SO4 are required per mole of NaOH.

However, Achiever's explanation presents a confusing way to think of it. Instead use the equation n1M1V1 = n2M2V2, which is essentially the same equation as your standard m1v1=m2v2 except that it takes into account polyprotic acids. The n represents the number of ionizable protons in this case (which is two for H2SO4).

2(3)(V1) = (1)(4)(30)
V1 = 20 mL

 

[J777]

Which is what rabbit said! (edited post as I typed up mine)

 

[dkidemo]

Thanks a lot. I guess you can't go wrong with the N1M1V1=N2M2V2 equation. Definitely going to store that in the memory bank for the exam. Am I also allowed to assume if it is H3PO4 and triprotic I would use 3 instead of 2 for N?

 

[Imazergling]

Thanks a lot. I guess you can't go wrong with the N1M1V1=N2M2V2 equation. Definitely going to store that in the memory bank for the exam. Am I also allowed to assume if it is H3PO4 and triprotic I would use 3 instead of 2 for N?

no

 

[AM4EVA]

no[/QU

why not? is it bcuz the third proton dissociated weakly and so is diprotic??

 

[J777]

I am pretty sure that it would apply for h3po4 also. Since there are 3 ionizable protons, n would = 3. There is a similar q to this one in the 2013 destroyer which deals with citric acid (also has 3 ionizable protons).. Question 212.