Gen Chem problem. Need help.

30.0 mL of 0.2 M of Ba(OH)2 is required to neutralize 25 ml of H3C6H5O7 . What is the molarity of the H3C6H5O7 ?
a. 0.167 M
b. 0.55 M
c. 0.25 M
d. 0.75 M
e. 0.084 M

not sure if this is the right answer, but here's how i'd do it:
you have 30.0 mL of .2 M Ba(OH)2, which is .030 L of .2 M, or .006 mol of Ba(OH)2. But since there's 2 moles of OH- in that, that's .012 mol of OH- needed to neutralize the H+ in the acid.
I'm assuming from the structure of the acid (with the 3 H in the front) that there are 3 H+ per molecule of acid, or 3 mole H+ per mole of acid. So if there's .012 mole of OH- to neutralize the H+, the acid has 1/3 that many moles.
.012/3 = .004 moles of acid / .025 L = .16 M, so I'd say A

hope this isn't embarrassingly super wrong!

 

yea what happy has is correct. You just have to remember to use the Normality equation and not Molarity because of the different # of equivalents. And side note, since they're both equal to each other, you don't really have to bother converting the mL to L.

 

wait but wut about the other 5 hydrogens, cuz its H3C6H5O7, isnt there 3+5 hydrogens or do you only count the terminal ones?

 

you only count the terminal ones. the ones in the middle are part of the molecule. they're not available for dissociating into solution.

 

because there are 2 equivalents of hydroxide ions that are in solution. Ba(OH)2 thus you multiply the molarity by 2. The same goes for H3C6H5O7. There are 3 protons that can dissociate into solution. That's why you divided .5 by 3. Just like you multiply the molarity of the Ba(OH)2 by 2, you multiply this one by 3. If you have kaplan, it goes into normality a little bit. Enough to understand i think.

 

no need to thank me. we're even now haha. you helped me understand titration! god this forum is a godsend...