Question

1. Assuming the density of a 5% by the mass acetic acid solution is 1.0 g/mL, determine the volume of acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH.

Answer 1

CH₃COOH + NaOH → CH₃COONa + H₂O

Proportion 1 (NaOH):

1000 mL – 0.10 mol

25.0 mL – n

n(NaOH) $= \frac{25.0 \times 0.10}{1000} = 0.0025 \;mol$

n(NaOH) = n(CH₃COOH) = 0.0025 mol

C(CH₃COOH) = 5 % or 5 g/100 mL or 50 g/L

M(CH₃COOH) = 180 g/mol

C(CH₃COOH) $= \frac{50 \;g/L}{180 \; g/mol} = 0.277 \;mol/L = 0.277 \;M$

Proportion 2:

0.277 mol – 1000 mL

0.0025 mol – x

$x = \frac{0.0025 \times 1000}{0.277} = 9.0\; mL$

Answer: 9.0 mL