[jbourne]

I've been having trouble with these problems and have no idea how to solve them:

1. Calculate the molarity of a sodium hydroxide if 133.75 mL of the solution is needed to neutralize 291 mL of 0.677 M oxalic acid. The equation for the reaction is:

H2C2O4 + 2NaOH ----> 2Na + C2O4

2. Calculate the volume of 2.19M sulfuric acid that would be needed to neutralize 46.4 mL of a 1.94 M aqueous ammonia solution. The equation for the reaction is:

H2SO4 + 2NH3 -----> 2NH4 (+)   + SO4  (-2)

Sorry for the formatting of the equations.  If anyone could take me step by step through solving these problems I would greatly appreciate it.

[sjb]

I've been having trouble with these problems and have no idea how to solve them:

1. Calculate the molarity of a sodium hydroxide if 133.75 mL of the solution is needed to neutralize 291 mL of 0.677 M oxalic acid. The equation for the reaction is:

H₂C₂O₄ + 2NaOH ----> 2Na⁺ + C₂O₄^2-

OK, what does the equation tell you about the ratio of the two compounds?

[jbourne]

I've been having trouble with these problems and have no idea how to solve them:

1. Calculate the molarity of a sodium hydroxide if 133.75 mL of the solution is needed to neutralize 291 mL of 0.677 M oxalic acid. The equation for the reaction is:

H₂C₂O₄ + 2NaOH ----> 2Na⁺ + C₂O₄^2-

OK, what does the equation tell you about the ratio of the two compounds?

1:1?

[Hunter2]

No, Think again.
How many sodium hydroxide is used according the equation (which is by the way not complete).

[Vidya]

Try to read the equation :
1 mole of the acid  is reacting with 2 moles of the NaOH so the ratio is 1:2
so you can say number of moles of the NaOH are double of number of moles of the oxalic acid.
moles of oxalic acid you can calculate from its molarity and volume used.