Question

I realize this is a general chemistry stoichiometry question, however I can not fully grasp it.

I am supposed to calculate the theoretical amount (volume) of 3M sodium hydroxide (NaOH) needed to convert 1.5g of benzoic acid to its salt.

I got 0.012 moles of bezoic acid present at the beginning and by looking at the balanced equation I see that there is a 1:1 ratio between the benzoic acid and NaOH, however I have no idea where to go from here. I am so confused.

Could someone help as soon as possible? Thanks!

Answer 1

You've done the hard part.
Yes, 1.5 g benzoic acid is 0.123 moles which means you must have 0.123 moles NaOH. What's the definition of molarity?
M = #moles/L; rearrange that to
M x L = #moles
So, you have # moles and M, calculate L and round to two significant figures. (I know 0.123 is more than that but I routinely carry one more place than I need and round at the end; otherwise, sometimes I make a rounding error.)
Here is a worked example of a stoichiometry problem that you may want to copy. It's a good step by step procedure.
http://www.jiskha.com/science/chemistry/stoichiometry.html

Answer 2

Thanks! and you do mean that it is 0.0123 moles, therefore I would plug in 3 for M, and 0.0123 for moles so that implies 3 x L = 0.0123 --> 0.0123/3 = L = .0041 L ~ 4.0mL ?

Answer 3

yes. that would be 4.1 mL of 3 M NaOH to add, theoretically, and if that is 3 M and not 3.0 M, then you may have only one s.f. and your 4 would be right (but not 4.0). If it is 3.0 M NaOH, I would round to 4.1 mL. Since benzoic acid can be titrated with NaOH, that really isn't a "highly" theoretical value but you may calculate exact amounts if the significant figures are there. I mean, in a quant lab, you could titrate to 4.12 mL or something like that. Of course, if it were a quant lab one wouldn't use 3 M NaOH to titrate it, either. :-).