Problem 82 - Chemistry: The Central Science

Question

A sample of \(0.2140 \mathrm{~g}\) of an unkown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Step-by-Step Solution

Answer

The molar mass of the unknown monoprotic acid is 82.2 g/mol, and its dissociation constant (Ka) is \(3.8 \times 10^{-5}\).

Step 1: Determine the amount of NaOH used at the equivalence point

To find the molar mass of the acid, we first need to determine the amount of NaOH used at the equivalence point. We can use the concentration and volume of NaOH given in the problem to calculate this: Amount of NaOH = concentration × volume Amount of NaOH = (0.0950 mol/L) × (27.4 mL × 1 L/1000 mL) Amount of NaOH = 0.002603 mol

Step 2: Calculate the moles of acid in the sample

Since the acid is monoprotic, we know that the moles of acid in the sample is equal to the amount of NaOH required to reach the equivalence point. Moles of acid = Amount of NaOH = 0.002603 mol

Step 3: Calculate the molar mass of the unknown acid

We can now find the molar mass of the unknown acid using the moles of acid and mass of the sample: Molar mass = mass of sample / moles of acid Molar mass = (0.2140 g) / (0.002603 mol) Molar mass = 82.2 g/mol (a) The molar mass of the acid is 82.2 g/mol.

Step 4: Determine the amount of NaOH added at pH 6.50

To find the Ka of the acid, we first need to know the amount of NaOH added at pH 6.50: Amount of NaOH at pH 6.50 = concentration × volume Amount of NaOH at pH 6.50 = (0.0950 mol/L) × (15.0 mL × 1 L/1000 mL) Amount of NaOH at pH 6.50 = 0.001425 mol

Step 5: Calculate the moles of acid and its conjugate base at pH 6.50

Since the total moles of acid remains constant, we can calculate the moles of acid and its conjugate base at pH 6.50: Moles of conjugate base = moles of NaOH added Moles of conjugate base = 0.001425 mol Moles of acid remaining = initial moles of acid - moles of NaOH added Moles of acid remaining = 0.002603 mol - 0.001425 mol Moles of acid remaining = 0.001178 mol

Step 6: Calculate the concentrations of the acid and its conjugate base at pH 6.50

The concentration of each species can be calculated by dividing the moles by the total volume of the solution at pH 6.50 (25.0 mL + 15.0 mL = 40.0 mL): Concentration of acid = moles of acid remaining / total volume Concentration of acid = (0.001178 mol) / (40.0 mL × 1 L/1000 mL) Concentration of acid = 0.02945 M Concentration of conjugate base = moles of conjugate base / total volume Concentration of conjugate base = (0.001425 mol) / (40.0 mL × 1 L/1000 mL) Concentration of conjugate base = 0.03563 M

Step 7: Calculate the dissociation constant Ka

Now we can use the given pH value and the concentrations of acid and conjugate base to find the Ka of the unknown acid. Since the pH = 6.50, the concentration of H+ ions is \(10^{-pH} = 10^{-6.50} = 3.16 \times 10^{-7}\,M\). Using the equilibrium expression for the dissociation of the acid (Ka = \([H^+][A^-]/[HA]\), we can solve for Ka: Ka = (3.16 × 10^{-7})(0.03563) / (0.02945) Ka = 3.8 × 10^{-5} (b) The Ka for the unknown acid is 3.8 × 10^{-5}.

Key Concepts

Molar Mass Calculation

Understanding how to calculate the molar mass of a substance is essential in chemistry. It involves determining the weight in grams of one mole of a compound. In the given exercise, we had a 0.2140 g sample of an unknown monoprotic acid, which was titrated with NaOH. To find the molar mass:

• First, measure the amount of NaOH used at the equivalence point: \[\text{Amount of NaOH} = 0.0950 \frac{\text{mol}}{\text{L}} \times 0.0274 \text{ L} = 0.002603 \text{ mol}\]
• This amount also equals the moles of the acid, since the acid is monoprotic.
• The molar mass is then found by dividing the sample mass by the moles of acid: \[\text{Molar Mass} = \frac{0.2140 \text{ g}}{0.002603 \text{ mol}} = 82.2 \text{ g/mol}\]

Calculating molar mass helps identify substances and understand their physical and chemical properties.

Acid-Base Equilibrium

Acid-base equilibrium concerns the reactions between acids and bases, where they neutralize each other. During titration, an acid like the unknown sample in the exercise, reacts with a base (NaOH), reaching a point where they've completely reacted—the equivalence point.

• The equivalence point tells us that moles of acid equal moles of the base added.
• This allows us to explore the relationship between acid concentration and its dissociation into ions.
• Titrations are used to determine various properties of an acid, like molar mass or dissociation constant.

Understanding this equilibrium helps in analyzing how much reactant is present and what reactions take place during the titration process.

Dissociation Constant (Ka)

The dissociation constant, \(K_a\), measures the strength of an acid in solution, depicting its ability to donate protons. It reflects the equilibrium position between the acid (HA), its conjugate base (A⁻), and protons (H⁺): \[K_a = \frac{[H^+][A^-]}{[HA]}\]

• During titration, the \(K_a\) value can be calculated using the concentrations of the acid and its conjugate base.
• At a given \(\text{pH} = 6.50\), find \([H^+]\) as \(10^{-6.50} = 3.16 \times 10^{-7} \text{ M}\).
• Use the calculated concentrations in the formula to find \(K_a\): \[K_a = \frac{(3.16 \times 10^{-7})(0.03563)}{0.02945} = 3.8 \times 10^{-5}\]

This \(K_a\) value indicates the relative strength of the acid and is a crucial parameter in assessing its behavior in various chemical environments.