What is the molarity of the potassium hydroxide if 25.25 ml of koh is required to neutralize 0.500 g of oxalic acid, h2c2o4? h2c2o4(aq)+2koh(aq)→k2c2o4(aq)+2h2o (l)

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) =

Now use the molar ratio to find how many moles of NaOH would be required to neutralize of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

: x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water: