Question

A solution is made by dissolving 0.40 g of solid sodium hydroxide in 250 cm3 of water. It is found that 28.2 cm3 of this solution is needed to completely neutralise a 40.0 cm3 sample of ethanoic acid (CH_3COOH ) of unknown concentration. Calculate the concentration of ethanoic acid​

Answer 1

Answer:

Molarity/concentration of ethanoic acid = 0.0282M

Explanation:

Amount of NaOh = 0.4 grams

Molar mass of NaOH = 40g/mol

No. of moles of NaOH = 0.4 / 40 = 0.01 moles

Volume of solution = 250 cm^3 = 0.25 liters      {1 liter = 1000 cm^3}

Molarity of NaOH (concentration) = 0.01 / 0.25

                                                        = 0.04 M

Volume of NaOH = 28.2 cm^3 = 0.0282 liters

Volume of ethanoic acid (CH3COOH) = 40 cm^3 = 0.04 liters

Molarity of ethanoic acid = M mol/L

MV (CH3COOH) = MV (NaOH)

0.04 M = 0.04 x 0.0282

M = 0.0282M