A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was ________ M. A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was ________ M. 0.175 0.119 0.365 0.263 none of the above
This is all i know 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
