Question

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?

H2C2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)

Answer 1

Concentration of KOH = 1.154 M

Step-by-step explanation:

`H_2C_2O_4(aq) + 2KOH(aq) \rightarrow K_2C_2O_4(aq) + 2H_2O(l)`

In the above reaction, 1 mole of oxalic acid reacts with 2 moles of KOH.

Mass of oxalic acid = 0.604 g

`Mole = (Mass\; in\;g)/(Molecular\;mass)`

Molecular mass of oxalic acid = 90.03 g/mol

`Mole = (0.604)/(90.03)=0.0067\;mol`

1 mol of oxalic acid reacts with 2 moles of KOH

0.0067 mol of oxalic acid reacts with
`0.0067* 2 = 0.0134 mol\; of\;KOH`

Volume of the solution = 27.02 mL = 0.0272 L

`Molarity=(Mole)/(Volume\;in\;L)`

No. of mole of KOH = 0.0134 mol

`Molarity=(0.0134)/(0.0272)=1.154\;M`

Concentration of KOH = 1.154 M