Question

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)

Answer 1

Answer:

0.444 mol/L

Explanation:

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

n(oxalic acid) = $\frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol$

Now use the molar ratio to find how many moles of NaOH would be required to neutralize $5.5537*10^{-3} mol$ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

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

$5.5537*10^{-3} mol$ : x

x = 0.0111 mol

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

$c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L$