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A 25.0mL sample of an acetic acid solution is titrated with a 0.175M NaOH solution. The equivalence point is reached when 37.5mL of the base is added. The concentration of acetic acid in the sample was :_________.A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119

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ysabayomi

Answer:

Explanation:

First of all, let's write the equation of the reaction.

CH₃COOH + NaOH  ⇒ CH₃COONa + H₂O

The formula to be used here is CaVa/CbVb = na/nb

where Ca is the concentration of the acid (?)

Cb is the concentration of the base (0.175M)

Va is the volume of the acid (25 mL)

Vb is the volume of the base (37.5 mL)

na is the number of moles of acid (1)

nb is the number of moles of base (1)

Ca × 25/0.175 × 37.5 = 1/1

Ca = 0.175 × 37.5 × 1 /25 ×1

Ca = 0.263 M

The concentration of the acid in the sample was 0.263 M

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