The molarity of the acetic acid, HC₂H₃O₂ solution is 0.244 M
We'll begin by writing the balanced equation for the reaction
HC₂H₃O₂ + KOH —> KC₂H₃O₂ + H₂O
From the balanced equation above,
The mole ratio of the acid, HC₂H₃O₂ (nA) = 1
The mole ratio of the base, KOH (nB) = 1
From the question given above, the following data were obtained:
Volume of the acid, HC₂H₃O₂ (Va) = 25 mL
Volume of the base, KOH (Vb) = 29.7 mL
Molarity of the base, KOH (Mb) = 0.205 M
Molarity of the acid, HC₂H₃O₂ (Ma) =?
MaVa / MbVb = nA/nB
(Ma × 25) / (0.205 × 29.7) = 1
(Ma × 25) / 6.0885 = 1
Cross multiply
Ma × 25 = 6.0885
Divide both side by 25
Ma = 6.0885 / 25
Ma = 0.244 M
Therefore, the molarity of the acetic acid, HC₂H₃O₂ solution is 0.244 M
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