Question

(a) If a titration of a different 10.0 mL sample requires 0.00500 moles of base, what mass of acetic acid is in the solution?

(b) Assuming the solution has a density of 1.0 g/mL, what is the mass % of acetic acid in the solution?

Answer 1

Answer :

a) # of moles of Base = 0.00500 moles.

Let, concentration of AcOH be 'M 'mole/L. Given volume of AcOH sample = V = 10 mL = 0.01 L.

# of moles of AcOH = Molarity x Volume in L = M x 0.01 = 0.01M moles.

At neutralization,

# of moles of AcOH = # of moles of Base

0.01M = 0.00500

M = 0.00500 / 0.01

M = 0.5 M

[AcOH] = 0.5 M

Moles of AcOH = moles of base = 0.00500.

Mass of AcOH = # of moles of AcOH x Molar mass of AcOH = 0.005 x 60.00 = 0.3 g.

Mass of AcOH in sample = 0.3 g.

b) Density of given AcOH solution is 1 g/mL.

So, Mass of 10 mL sample = Volume x density = 10 x 1 = 10 g.

Now,

% Mass of AcOH = (Mass of AcOH / Mass of Sample) x 100

= (0.3/10) x 100

= 3 %.

Mass of AcOH in sample = 3 %.

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