[ohiohelp]

Full question is below.

I'm so lost you guys. And I'm frustrated beyond belief. If anybody has any idea of how to do this PLEASE SHARE! I think this has something to do with titration but I thought you had to know BOTH volumes not just one. I was only given one. See below. 

thanks!

Again... In the equation HNO2 + NaOH ----------> H20 + NaNO3

28mL of 0.500 M HNO3 is neutralized by a 0.300 M solution of NaOH. How much NaOH solution is used.

[ARGOS++]

Dear ohiohelp;

You are asked to calculate this second volume (volume of used NaOH).
You know all what is required and your RxN is ok. Yes it is titration!

Good Luck!
                    ARGOS⁺⁺

[ohiohelp]

Thank you !! Okay can someone tell me if I'm still on the right track

I took .028L/1L x .500M = .014M for the concentration

Is this right...but then what do I do or is that the answer lol. I'm so lost 

[ohiohelp]

Okay my final answer...and I know I'm wrong with the decimal place...

.021L

I just took .300 / .014 = x

I actually got 21.42 but I know that can be L ...once again any feedback is greatly appreciated 

[Borek]

http://www.titrations.info/titration-calculation

You have to calculate number of moles of acid first (hint: molarity definition will came handy). Then you have to calculate number of moles of base (hint: reaction equation tells you ratio of the reaction). Finally, you have to calculate what volume contains required number of moles of base (hint: again, molarity definition).

And pay attention to units, as you have to convert between mL and L.