Problem 36 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A \(0.28 \mathrm{~g}\) sample of an unknown monoprotic organic acid is dissolved in water and titrated with a 0.1 M sodium hydroxide solution. After the addition of \(17.5 \mathrm{ml}\) of base, a pH of \(5.0\) is recorded. The equivalence point is reached when a total of \(35.0 \mathrm{ml}\) of \(\mathrm{NaOH}\) is added. The molar mass of the organic acid is (a) 160 (b) 80 (c) 40 (d) 120

Step-by-Step Solution

Answer

The molar mass of the organic acid is (b) 80 g/mol.

Step 1: Determine moles of NaOH at equivalence point

At the equivalence point, the amount of NaOH added is stoichiometrically equivalent to the amount of the unknown acid in the solution. First, calculate the moles of NaOH using the volume at the equivalence point and its molarity: \( Moles_{NaOH} = Molarity_{NaOH} \times Volume_{NaOH} \)\( Moles_{NaOH} = 0.1~M \times 0.0350~L = 0.0035~moles \)

Step 2: Determine the molar mass of the acid

Since the acid is monoprotic, one mole of acid reacts with one mole of NaOH. The moles of acid is equal to the moles of NaOH at equivalence point. Using the mass of the acid sample, calculate its molar mass:\( Molar Mass_{acid} = \frac{Mass_{acid}}{Moles_{acid}} \)\( Molar Mass_{acid} = \frac{0.28~g}{0.0035~moles} = 80~g/mol \)

Step 3: Choose the correct answer

With the calculated molar mass of 80 g/mol, select the appropriate answer among the choices given.

Key Concepts

Understanding Monoprotic Acids

Monoprotic acids are a key concept in acid-base chemistry. They are defined as acids that donate one proton (hydrogen ion) per molecule during a chemical reaction. The proton donation typically occurs when the acid reacts with a base in a neutralization reaction. Water (H₂O), being the universal solvent, often serves as the medium for these reactions.

For instance, the unknown organic acid in our exercise is monoprotic, which means for every molecule of the acid that dissolves in water, it releases one hydrogen ion. This is crucial for calculating stoichiometry in titration problems, as we will see next. In a titration with sodium hydroxide (NaOH), a strong base, each NaOH molecule will neutralize one hydrogen ion from the monoprotic acid, forming water and a salt.

Decoding the Equivalence Point

The equivalence point of a titration is a fundamental concept that denotes the exact point at which the amount of titrant (in this case, NaOH) added is stoichiometrically equivalent to the quantity of the substance being titrated (the unknown monoprotic acid).

At the equivalence point, the number of moles of base is equal to the number of moles of the monoprotic acid. This is vital information for determining the molar mass of the acid. In our exercise, the equivalence point is achieved when 35.0 ml of 0.1 M NaOH is added. The pH at this stage is not given or required since we know the stoichiometry of the reaction is one-to-one; every mole of acid reacts with a mole of base.

Calculating Molar Mass

Molar mass calculation is an integral part of solving titration problems when you need to determine the molecular weight of an unknown substance. The molar mass is the weight in grams of one mole of a substance.

After determining the moles of NaOH used at the equivalence point, we infer that this is also the moles of our unknown acid due to the one-to-one reaction ratio. In our problem, the sample weight of the acid (0.28 grams) divided by the moles (0.0035 moles) gives us the molar mass of the acid—80 grams per mole. This process allows us to discern the identity or purity of an unknown sample by comparing the calculated molar mass to known values.