Answer to Question #326271 in General Chemistry for Francis

Question

The titration of a 10.0ml sample of vinegar requires 30.0ml of 0.20 M NaOH solution. Calculate

i. The molarity

ii. The mass /mass percent concentration of acetic acid.

Given that the density of acetic acid is 1.01g/cm3

Answer 1

V(CH₃COOH) = 10.0 mL = 0.010 L;

V(NaOH) = 30.0 mL = 0.030 L;

C(NaOH) = 0.20 M;

CH₃COOH + NaOH = CH₃COONa + H₂O;

By the chemical reacion:

C(CH₃COOH) * V(CH₃COOH) = C(NaOH) * V(NaOH);

C(CH₃COOH) = C(NaOH) * V(NaOH)/V(CH₃COOH) = 0.20 * 0.030/0.010 = 0.60 M.

n(CH₃COOH) = C(NaOH) * V(NaOH) = 0.20 * 0.030 = 0.006 mol;

M(CH₃COOH) = 60 g/mol;

m(CH₃COOH) = n(CH₃COOH) * M(CH₃COOH) = 0.006 * 60 = 0.36 g;

d(vinegar) = 1.01 g/sm³ = 1.01 g/mL;

m(vinegar) = d(vinegar) * V(vinegar) = 1.01 * 10 = 10.10 g;

X(CH₃COOH) = m(CH₃COOH)/m(vinegar) = 0.36/10.10 = 0.0356 = 3.56 %.

Answer: The molarity is 0.60 M;

The mass /mass percent concentration of acetic acid is 3.56 %.