Question

(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50 mL of titrant, how many mmol of NaOH are required to reach the endpoint?

__?__mmol NaOH

(b) How many mmol of acetic acid (HC₂H₃O₂) are required to react with the NaOH?
__?__ mmol HC₂H₃O₂

(c) How many grams of acetic acid is this?
__?__ grams HC₂H₃O₂

(d) If the mass of analyte is 10.05 grams, what is the mass % of acetic acid in the analyte?
__?__ %

Answer 1

a)

mmol of NaOH = M(NaOH) * Volume in mL

= 0.410 M * 13.50 mL

= 5.535 mmol

Answer: 5.54 mmol

b)

mol of HC2H3O2 = mol of NaoH

So,

mmol of HC2H3O2 = 5.535 mmol

Answer: 5.54 mmol

c)

Molar mass of HC2H3O2,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of HC2H3O2,

m = number of mol * molar mass

= 5.535*10^-3 mol * 60.05 g/mol

= 0.3324 g

Answer: 0.332 g

d)

mass % = mass of HC2H3O2 * 100 / total mass

= 0.332 * 100 / 10.05

= 3.30 %

Answer: 3.30 %