To determine the molecular weights of the acid, a student dissolved a pure sample (400.0 mg) in a water to a total volume of 100.0 mL. The student then took a 25.00 mL aliquot of this solution and titrated it with 15.90 mL 0.0750 M NaOH to reach a phenol red end point. The molecular weight of the acid is: (show your work)

83.86g/mol is the molecular weight of the acid

Explanation:

Molecular weight is the ratio of mass in grams and moles present. That means we need to determine the mass in grams (400.0mg = 0.400g) and the moles of the acid that can be detemined by using its reaction with NaOH:

Moles acid:

First, we need to assume the acid is a monoprotic acid (Only 1 proton is acid and reacts with a base).

when the acid, HX, reacts with NaOH there is produced:

HX + NaOH → H₂O + NaX

Where 1 mole of HX reacts per mole of NaOH. That means the moles at the endpoint of NaOH = Moles of HX:

Moles NaOH:

15.90mL= 0.01590L * (0.0750mol / L) = 1.193x10⁻³ moles of NaOH = Moles of HX are present.

The acidic solution was diluted to 100.0mL but only 25.00mL were taken, that is:

25.00mL / 100.00mL = 1/4 of the acid was titrated.

And moles of HX in the original solution are:

1.193x10⁻³ moles of HX * 4 = 4.77x10⁻³ moles of HX

It is multiplied by 4 because the real amount of acid is 4 times the titrated amount.

Thus, molecular weight is:

0.4000g / 4.77x10⁻³ moles of HX =

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