Answer to Question #327446 in General Chemistry for Emy

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

The titration reaction of the acetic acid by the sodium hydroxide:

$CH_3COOH + NaOH = CH_3COONa + H_2O$

i. The molarity of acetic acid titrated can be deduced from the following equation:

$n(CH_3COOH)=n(NaOH)$

$c(CH_3COOH)\times V(CH_3COOH) = c(NaOH) \times V(NaOH)$

$c(CH_3COOH)=\frac{c(NaOH) \times V(NaOH)}{V(CH_3COOH)}=\frac{0.20M \times 30.0 ml}{10.0 ml} = 0.60M$

ii. Mass of the solute is determining by the formula:

$c(CH_3COOH)=\frac{n(CH_3COOH)}{V} = \frac{m \times M}{V}$

$m(CH_3COOH)=\frac{c\times V}{M}=\frac{0.6 mol/L \times 0.01L}{60.05 g/mol}=0.0001g=10^{-5}g$

Mass of the solution:

$m = V\times d = 10.0 ml \times 1.01 g/cm^3 = 10.1 g$

Mass/mass percent concentration of acetic acid:

$W = \frac{m(CH_3COOH)}{m} \times 100 \% = \frac {0.0001 g}{10.1 g}\times 100 = 0.00099 \%$