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

0.444 mol/L

Step-by-step explanation:

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

n(oxalic acid) =
`(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 = (number of moles)/(volume) = (0.0111)/(0.025) = 0.444 mol/L`