# GC question

#### ferfax

If 50 mL of H2SO4 required 50.0 mL of 1.0 M NaOH for a complete neutralization reaction, according to the equation shown below. What would be the molarity of an acid?

2NaOH + H2SO4 -> Na2SO4 + 2 H2O

I started off by using the n1M1V1=n2M2V2 equation but whats really tripping me up is the balanced equation they showed below? Why is that? Would it alter me using that formula at all?

D

#### deleted982220

This is a titration, you are trying to figure out the original concentration of the acid. This is how you do it according to the balanced reaction.
0.05 L X 1.0 moles/L = 0.05 moles NaOH
0.05 moles NaOH X 1 moles H2SO4/2 moles NaOH= 0.025 moles H2SO4 present in the original acid solution
0.025 moles H2SO4/ 0.05L = 0.5 M for the original acidic solution

I would recommend watching a video of an actual laboratory titration and the goal of a titration, doing that helped me with these types of problems.

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D

#### deleted982220

For N1V1=N2V2, there is no need for the balanced reaction. This is why this equation is useful, no need to do all that I did above.
1 M NaOH, N= 1X1M=1N; V= 0.05L
H2SO4, N2= 2XM; V2=0.05

N1V1=N2V2
(1N)(0.05)=N2(0.05)
N2=1=2XM
M=0.5

Let me know if you need further clarification, I'd be happy to help.

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#### ferfax

For N1V1=N2V2, there is no need for the balanced reaction. This is why this equation is useful, no need to do all that I did above.
1 M NaOH, N= 1X1M=1N; V= 0.05L
H2SO4, N2= 2XM; V2=0.05

N1V1=N2V2
(1N)(0.05)=N2(0.05)
N2=1=2XM
M=0.5

Let me know if you need further clarification, I'd be happy to help.
Thank you for the help

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