Molarity ASAP HELP!!

Question

molarity of the KOH solution used when 10 mL of a 0.121 M H2SO4 solution is neutralized by 17.1 mL of the KOH solution

H2SO4 2KOH -> 2H2O K2SO4

this is what I got-is it right?

10mlx2x.121)/17.1mL = .14

Answer 1

Thats correct.  Thats why its better to work in normality, to avoid the s***w up of knowing that there are 2 hydroniums per mol of H2SO4.

Answer 2

Moles H2SO4 = 0.010 L x 0.121 =0.00121

the ratio between H2SO4 and KOH is 1 : 2

Moles KOH needed = 2 x 0.00121 = 0.00242

M of KOH = 0.00342 / 0.0171 L = 0.142

Answer 3

Start with the balanced chemical equation:

H2SO4(aq) + 2KOH(aq) --> K2SO4(aq) + 2HOH(l)

Starting with the moles of H2SO4, compute the molarity of the KOH using the unit factor method (conversion factors).  Using the unit factor method and canceling out the units we don't want, while leaving the units that we do, will insure me of getting the correct "set up" for the problem.

10.0 mL x (0.121 mol H2SO4 / L) x (2 mol KOH / 1 mol H2SO4) x (1 / 17.1 mL) = 0.142 M KOH

You got the correct answer, but your setup leaves something to be desired.  In my classroom, you would have lost some points, even though you got the correct answer. You didn't adequately show your work.  You didn't show any units, even with the answer, and you need to pay attention to the number of significant digits, so that the answer will have the correct precision.  It will all depend on how little your teacher will let you get by with.

===== Follow up ======

I'm curious to know why using the stoichiometric relationship provided by the balanced chemical equation is a "s***w up".

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58.5 g of NaCl dissolved in 0.50 L of solution

This one is easy enough to do in our heads.  58.5 g is the molar mass of NaCl, so it is 1.00 mol. Therefore 1.00 mol of NaCl is dissolved in half a liter of solution.  In order to have the same ratio, 2.00 moles must be dissolved in 1.00 L.  And there you have the concentration: 2.0 mol / L or 2.0M.