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What is the molarity of a NaOH solution when 22.14 mL of 0.105 M oxalic acid is needed to
neutralize 25.0 mL of the base.

Sagot :

Eduard22sly

The molarity of the NaOH solution required for the reaction is 0.186 M

We'll begin by writing the balanced equation for the reaction

H₂C₂O₄ +2NaOH —> Na₂C₂O₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂C₂O₄ (nA) = 1

The mole ratio of the base, NaOH (nB) = 2

From the question given above, the following data were obtained:

Volume of the acid, H₂C₂O₄ (Va) = 22.14 mL

Molarity of the acid, H₂C₂O₄ (Ma) = 0.105 M

Volume of the base, NaOH (Vb) = 25 mL

Molarity of the base, NaOH (Mb) =?

MaVa / MbVb = nA/nB

(0.105 × 22.14) / (Mb × 25) = 1/2

2.3247 / (Mb × 25) = 1/2

Cross multiply

Mb × 25 = 2.3247 × 2

Mb × 25 = 4.6494

Divide both side by 25

Mb = 4.6494 / 25

Mb = 0.186 M

Therefore, the molarity of the base, NaOH solution is 0.186 M

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