Question

You used 1.512 g of your unknown weak acid and it took 37.98 mL of your standardized 0.107 M NaOH to reach the 2nd equivalence point, using the procedure described in the lab manual. What is the molecular weight of the acid

Answer 1

The molecular weight of the unknown weak acid is calculated to be 744.1 g/mol, using the number of moles of NaOH at the second equivalence point and the mass of the weak acid used in the titration.

Step-by-step explanation:

To determine the molecular weight of an unknown weak acid from a titration experiment, we first need to find the number of moles of NaOH used to reach the second equivalence point. Since the concentration of the NaOH solution is given as 0.107 M and the volume used is 37.98 mL (which we convert to liters by dividing by 1000), we can calculate the moles of NaOH:

Number of moles of NaOH = Volume of NaOH in liters × Concentration of NaOH

Number of moles of NaOH = 0.03798 L × 0.107 M = 0.00406146 mol

At the second equivalence point, the moles of weak acid are equal to half the moles of NaOH used, because two moles of NaOH would be necessary to neutralize one mole of the acid (considering a diacidic weak acid). Thus:

Moles of acid = 0.00406146 mol × 1/2 = 0.00203073 mol

With the mass of the weak acid (1.512 g) and the moles (0.00203073 mol), we can calculate the molecular weight:

Molecular Weight = Mass of acid × Moles of acid

Molecular Weight = 1.512 g / 0.00203073 mol = 744.1 g/mol

Answer 2

Step-by-step explanation:

37.98 mL of .107 M NaOH = 37.98 x .107 mL of M NaOH

= 4.064 mL of M NaOH .

= 4.064 x 10⁻³ L of M NaOH

It will contain 4.064 x 10⁻³ moles of NaOH

4.064 x 10⁻³ moles of NaOH will react with 4.064 x 10⁻³ moles of unknown acid

1.512 g of acid = 1.512 / M moles of acid where M is molecular weight of acid

1.512 / M = 4.064 x 10⁻³

M = 1.512 / 4.064 x 10⁻³

= 372 .

Mol weight of acid = 372.