Question

Lab Section Name Date Prelaborator laboratory Problems - Experiment 3-Acids and Bases: Analysis 1.- The ti Hesa, + 2004 the titration of 25.0 mL of a sulfuric acid solution of unknown concentration requires selo ml of a 0.1234 MNOH solution. What is the concentration of the sulfuric acid 0.1234 U Wa Ohm 0.03224 Na Son +2460 3.859548X10" mokes Da OH 2. 10.00 mL of vinegar (mass10.05 g) requires 16.20 end point. Calculate the molarity and mass per 0.05 g) requires 16.28 mL of 0.5120 M NaOH to reach the molarity and mass percent of the acetic acid in the vinegar. . 0.5120 MNOOH m 0011282 8.3 x 10° moles Na CH. Ligh A 0.1936 g sample of an unknown monoprotic acid requires 15.56 mL of 0.1020 M NOE solution to reach the end point. What is the molecular mass of the acid?

Answer 1

1) at neutralization point

mmol of H+ = mmol of OH-

M × 2×25.00 = 0.1234×31.22

M = 0.0770 M

So molality of sulfuric acid = 0.0770 M

2) At end point

Moles of Acetic Acid = moles of NaOH

Mass/molar mass = M × V

Mass/60 = 0.5120×16.28/1000

Mass = 0.50 gram

Moles of Acetic acid = 0.5/60 => 0.008335 mol

Molarity = 0.008334/0.01 = 0.8335 M

%mass=0.50×100/10.05 => 4.98%

3) At end point

Moles of Acid = moles of NaOH

0.1936/molar mass = 0.1020×15.56/1000

Molar mass = 122.0 g/mol

So, molar mass of monoprotic acid = 122.0 g/mol