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For a titration Maria added 38.4 mL of 0.250 M H2SO4 to 23.5
mL of K2CO3 solution to reach the equivalence point.
Given the reaction above, what is the molarity of the original
K2CO3 solution?

Answer :

Answer:

0.409 m

Explanation:

Eduard22slygo

The molarity of the original K₂CO₃ solution given the data from the question is 0.409 M

Balanced equation

H₂SO₄ + K₂CO₃ —> K₂SO₄ + CO₂ + H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₂SO₄ (nA) = 1
  • The mole ratio of the base, K₂CO₃ (nB) = 1

How to determine the molarity of K₂CO₃

  • Volume of acid, H₂SO₄ (Va) = 38.4 mL
  • Molarity of acid, H₂SO₄ (Ma) = 0.25 M
  • Volume of base, K₂CO₃ (Vb) = 23.5 mL
  • Molarity of base, K₂CO₃ (Cb) = ?

MaVa / MbVb = nA / nB

(0.25 × 38.4) / (Mb × 23.5) = 1

9.6 / (Mb × 23.5) = 1

Cross multiply

Mb × 23.5 = 9.6

Divide both side by 23.5

Mb = 9.7 / 23.5

Mb = 0.409 M

Thus, the molarity of the K₂CO₃ solution is 0.409 M

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