The molar concentration of the CH₃COOH soluttion is 0.33 M
We'll begin by writing the balanced equation for the reaction. This is given below:
CH₃COOH + NaOH —>CH₃COONa + H₂O
From the balanced equation above,
The mole ratio of the acid, CH₃COOH (nA) = 1
The mole ratio of the base, NaOH (nB) = 1
From the question given above, the following data were obtained:
Volume of acid, CH₃COOH (Va) = 15 mL
Molarity of base, NaOH (Mb) = 0.2 M
Volume of base, NaOH (Vb) = 24.55 mL
Molarity of acid, CH₃COOH (Ma) =?
MaVa / MbVb = nA/nB
(Ma × 15) / (0.2 × 24.55) = 1
(Ma × 15) / 4.91 = 1
Cross multiply
Ma × 15 = 4.91
Divide both side by 15
Ma = 4.91 / 15
Ma = 0.33 M
Therefore, the molar concentration of the acid, CH₃COOH is 0.33 M
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