What Is the Molarity of a Sulfuric Acid Solution Neutralized by NaOH?

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To find the molarity of a sulfuric acid solution neutralized by NaOH, the balanced chemical equation shows that two moles of NaOH react with one mole of H2SO4. Given that 20 mL of 0.3M NaOH is used, this equates to 0.006 moles of NaOH, which corresponds to 0.003 moles of H2SO4 due to the stoichiometric ratio. Dividing the moles of H2SO4 by the volume of the acid solution (0.030 L) yields a molarity of 0.1M for the sulfuric acid. The discussion emphasizes the importance of understanding the relationships in the chemical equations and breaking down the problem into manageable steps. Overall, the process illustrates a straightforward application of stoichiometry in acid-base neutralization.
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This is the question: If 20 mL of a 0.3M solution of NaOH is required to neutralize 30.0 mL of a sulfuric acid solution, what is the molarity of the acid solution?

I just need to get pointed in the right direction and then just some simple steps from there...

Thanks
H20h
 
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Do you mean like this:

2NaOH + H2SO4 ==> Na2SO4 + H2O ?
 
I am not sure? is that where I should start?
 
h20h said:
I am not sure? is that where I should start?
If you had to ask, then I would say, "probably so" (meaning I believe, "yes"). Your level of understanding seems to be near introductory, so you are interested in complete neutralization. You are interested in neutralizing both equivalents of acid. Use ratios from the balanced reaction equation. This means that one formula unit of sulfuric acid matches two formula units of sodium hydroxide.

If M is molarity, and V is volume, then M*V for sodium hydroxide will equal M*V for sulfuric acid. M_b V_b = 2 M_a V_a

(another attempt at typesetting; I hope it works)
 
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I am taking Chem 2 in college right now but I HATE word problems I never really know where to start...so using the equation you have above I would just plug in the values that were given and then rearrange to find the missing component?

I went from liters of NaOH to moles of NaOH =.006 moles of NaOH

stoich: 2 moles of NaOH: 1 mole of H2SO4

moles of NaOH to moles of H2SO4= .003 moles of H2SO4

then divided the moles of H2SO4 by .03 liters and got = .1M H2SO4 solution
 
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...Looks good, h20h. You broke the description into smaller steps.

That word problem as you presented it, is not very complicated in its worded form. Always identify the important mathematical relationships and laws; write equations and expressions for what you know; analyze the relationships and arrange an answer in the form of symbols.

What is the exact name of the course which you call, "Chem 2"?
 
It's just called CHEM 2 (I am not lying here LOL)

Thanks so much for your help I really appreciate it alot
 
h20h said:
It's just called CHEM 2 (I am not lying here LOL)

Thanks so much for your help I really appreciate it alot
h20h,
The course "Chem 2" must have an official descriptive title. "Chem 2" alone is not enough. I really AM interested to know what the course name is. Check the college catalogue; check the course schedule for this current term. Your Chem 2 course might not be a typical degree credit course, but still an official name is expected. With the kind of question you asked, the correct descriptive name might me something like "Introductory Chemistry", or "Elementary Chemistry"; but probably not "analytical chemistry" and not "quantitative chemistry" and not "principles of chemistry" and certainly not "General Chemistry". Could you give a list of topics for the course, or a syllabus?
 
Actually, I would guess that it is general chem. 2 that h2oh is in. I don't think they mess with an intro. to chem. 2.
However, the problem does seem a bit easy for gen. chem. 2, but after a bunch of hard classes, maybe it's a question of perspective.
 
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