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If 40mL of 2.50M HCl neutralizes 500ml. of KOH, what is the molarity of the KOH?

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Eduard22sly

The molarity of the 500 mL KOH solution needed to neutralize the 40 mL of 2.50 M HCl is 0.2 M

Balanced equation

HCl + KOH —> KCl + H₂O

From the balanced equation above,

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

The mole ratio of the base, KOH (nB) = 1

How to determine the molarity of KOH

  • Volume of acid, HCl (Va) = 40 mL
  • Molarity of acid, HCl (Ma) = 2.5 M
  • Volume of base, KOH (Vb) = 500 mL
  • Molarity of base, KOH (Mb) =?

MaVa / MbVb = nA / nB

(2.5 × 40) / (0.15 × Vb) = 1

100 / (Mb × 500) = 1

Cross multiply

Mb × 500 = 100

Divide both side by 500

Mb = 100 / 500

Mb = 0.2 M

Thus, the molarity of the KOH solution needed is 0.2 M

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