Question

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

Answer 1

= 0.2625M ≅ 0.26M (2 sig figs)

Step-by-step explanation:

HOAc + NaOH => NaOAc + H₂O

25ml (Molarity of HOAc) =\ 37.5ml(0.175M NaOH)

Molarity of HOAc = 37.5ml(0.175M NaOH)/25ml

= 0.2625M ≅ 0.26M (2 sig figs)

Answer 2

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