BR Gen Chem Section V (Buffers & Titrations)...

[BestDoctorEver]

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On passage VIII, question 50, I believe the answer should be A (0.125 M NaOH) instead of C (0.250 M NaOH). I dont understand why they're using 0.2 M sulfuric acid to calculate its mole. The passage gives the Normality of H2SO4 which is 0.2 N, not the Concentration. Therefore, if the Normality of H2SO4 is 0.2 N, the concentration should be 0.1 M. Consequently, they should use 0.1 M to calculate the mole of H2SO4. Can someone tell what I am missing in this question?

 

[BestDoctorEver]

On passage VIII, question 50, I believe the answer should be A (0.125 M NaOH) instead of C (0.250 M NaOH). I dont understand why they're using 0.2 M sulfuric acid to calculate its mole. The passage gives the Normality of H2SO4 which is 0.2 N, not the Concentration. Therefore, if the Normality of H2SO4 is 0.2 N, the concentration should be 0.1 M. Consequently, they should use 0.1 M to calculate the mole of H2SO4. Can someone tell what I am missing in this question?

bump

 

[Catburr]

Sulfuric acid is diprotic; you get 2 moles of H+ for every 1 mole of the acid in solution.

Sodium hydroxide gives just the 1 mole of OH- for every 1 mole of sodium hydroxide in solution, though.

They're not using 0.2M sulfuric acid to calculate NaOH needed, they're using 0.2M H+. You're exactly right that 0.2N sulfuric acid and 0.1M sulfuric acid are one and the same solution.

But you don't use that 0.1M sulfuric acid concentration to directly figure out how much NaOH you need, because then the NaOH gives one OH-, and the sulfuric acid gives two H+

Mathed out:
So because you have 25mL of 0.2M H+, thats 0.005 moles of H+ that we need to neutralize. How do you get the 0.005 moles of OH- that we need to neutralize that? 20mL of 0.25M NaOH will do it, because that's 20mL of 0.25M OH-.

I hope that makes sense, probably not the best possible explanation on my part!

 

[BestDoctorEver]

Sulfuric acid is diprotic; you get 2 moles of H+ for every 1 mole of the acid in solution.

Sodium hydroxide gives just the 1 mole of OH- for every 1 mole of sodium hydroxide in solution, though.

They're not using 0.2M sulfuric acid to calculate NaOH needed, they're using 0.2M H+. You're exactly right that 0.2N sulfuric acid and 0.1M sulfuric acid are one and the same solution.

But you don't use that 0.1M sulfuric acid concentration to directly figure out how much NaOH you need, because then the NaOH gives one OH-, and the sulfuric acid gives two H+

Mathed out:
So because you have 25mL of 0.2M H+, thats 0.005 moles of H+ that we need to neutralize. How do you get the 0.005 moles of OH- that we need to neutralize that? 20mL of 0.25M NaOH will do it, because that's 20mL of 0.25M OH-.

I hope that makes sense, probably not the best possible explanation on my part!

Explanation is perfect...It makes sense now. Thanks.