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

Answer:

Concentration of KOH = 1.154 M

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

$Mole = \frac{Mass\; in\;g}{Molecular\;mass}$

Molecular mass of oxalic acid = 90.03 g/mol

$Mole = \frac{0.604}{90.03}=0.0067\;mol$

1 mol of oxalic acid reacts with 2 moles of KOH

0.0067 mol of oxalic acid reacts with $0.0067\times 2 = 0.0134 mol\; of\;KOH$

Volume of the solution = 27.02 mL = 0.0272 L

$Molarity=\frac{Mole}{Volume\;in\;L}$

No. of mole of KOH = 0.0134 mol

$Molarity=\frac{0.0134}{0.0272}=1.154\;M$

Concentration of KOH = 1.154 M