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Answer :
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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