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Elan Coil [88]
3 years ago
15

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?

Chemistry
1 answer:
kakasveta [241]3 years ago
6 0

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

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