Question

A \(10.00-\mathrm{g}\) sample of the ionic compound \(\mathrm{NaA},\) where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution and was then titrated with \(0.100 \mathrm{M}\) HCl. After \(500.0 \mathrm{~mL}\) HCl was added, the \(\mathrm{pH}\) was 5.00 . The experimenter found that \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.

Short answer

a. The molar mass of NaA is \(100.00 \mathrm{g/mol}\). b. The \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration is approximately \(1.05\).

Step-by-step solution

Calculate the moles of HCl at the stoichiometric point

The volume of \(0.100 \mathrm{M}\) HCl required to reach the stoichiometric point is \(1.00 \mathrm{L}\). We can use the concentration and volume to find the moles of HCl: Moles of HCl = (Concentration of HCl) x (Volume of HCl) Moles of HCl = \(0.100 \mathrm{M} \times 1.00 \mathrm{L} = 0.100 \mathrm{mol}\)

Find the moles of A

At the stoichiometric point, the moles of HCl and A- are equal. Therefore, the moles of A- = moles of HCl = \(0.100 \mathrm{mol}\).

Calculate the moles of Na+

Since NaA dissociates into one Na+ and one A- ion, the moles of Na+ = moles of A- = \(0.100 \mathrm{mol}\).

Determine the molar mass of NaA

To find the molar mass of NaA, divide the mass of the sample by the moles of Na+: Molar mass of NaA = \(\frac{10.00 \mathrm{g}}{0.100 \mathrm{mol}} = 100.00 \mathrm{g/mol}\)

Determine the \(\mathrm{pH}\) of the solution at the stoichiometric point

At the stoichiometric point, the amounts of HCl and A− are equal, such that all the A− is converted into HA (the weak acid). Following the stoichiometric point, there is an excess of HCl in the solution. Since the \(\mathrm{pH}\) of the acidic solution depends on the \(\mathrm{pH}\) of the strong acid (HCl) in the solution, we can calculate the concentration of HCl at the stoichiometric point: Concentration of excess HCl = \(\frac{0.100 \mathrm{mol} \times (1.00 \mathrm{L} - 0.100 \mathrm{L})}{0.100 \mathrm{L}} = 0.0900 \mathrm{M}\) Now, we can find the \(\mathrm{pH}\) using the concentration of HCl: \(\mathrm{pH}\) = \(-\log (0.0900) \approx 1.05\) The \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration is approximately 1.05. In summary: a. The molar mass of NaA is \(100.00 \mathrm{g/mol}\). b. The \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration is approximately \(1.05\).

Key concepts

Molar Mass Calculation

Understanding molar mass calculation is crucial in chemistry for determining the amount of a substance. The molar mass is the weight of one mole of a chemical compound. It is expressed in grams per mole (g/mol) and can be calculated by dividing the mass of the sample by the number of moles it contains. In the case of the ionic compound NaA from the exercise, the sample mass is given as 10.00 g. The task is to find the moles, which is done by titration with HCl. At the stoichiometric point, equal moles of NaA and HCl are present, allowing us to know that we have 0.100 moles based on the titration data. Using the formula:\[\text{Molar Mass} = \frac{\text{Mass of Sample}}{\text{Moles of Sample}}\]we calculate:\[\text{Molar Mass of NaA} = \frac{10.00 \text{ g}}{0.100 \text{ mol}} = 100.00 \text{ g/mol}\]Thus, the molar mass of NaA is 100.00 g/mol.

pH Calculation

The pH calculation is key when working with solutions, as it indicates how acidic or basic the solution is. The pH scale ranges from 0 to 14, with 7 being neutral. Acidic solutions have pH values less than 7. The formula used to calculate pH, especially when dealing with strong acids like HCl, is:\[\mathrm{pH} = -\log [H^+]\]At the stoichiometric point in this scenario, any remaining amount of HCl after neutralization impacts the solution’s pH. The concentration of the excess HCl can be calculated from the given data:\[\text{Concentration of excess HCl} = \frac{\text{Initial moles of HCl} - \text{Moles needed to reach stoichiometric point}}{\text{Total volume}}\]\[\mathrm{Concentration} = \frac{0.100 \text{ mol} \times (1.00 \text{ L} - 0.100 \text{ L})}{0.100 \text{ L}} = 0.0900 \mathrm{M}\]With this concentration, the pH is computed as:\[\mathrm{pH} = -\log (0.0900) \approx 1.05\]Thus, the pH at the stoichiometric point is approximately 1.05.

Stoichiometric Point

The stoichiometric point, also known as the equivalence point in a titration, is the moment when the reactants have reacted completely in stoichiometric ratios according to the balanced chemical equation. For acid-base reactions, this means that the quantity of acid equals the quantity of base. In the given exercise, 1.00 L of 0.100 M HCl was reacted with the NaA solution. At this point, the moles of HCl added are equal to the moles of the anion (A−) of the weak acid from NaA. Hence, at the stoichiometric point:

• Moles of HCl = Moles of A− = 0.100 mol.
• No reactants remaining in excess from the original pairs reacting with each other.

Reaching this point is critical for accurately determining concentrations and performing subsequent calculations in titration experiments.

Ionic Compound

Ionic compounds consist of positively charged ions (cations) and negatively charged ions (anions) held together by ionic bonds. These compounds generally form through the transfer of electrons from a metal atom to a non-metal atom, leading to electrostatic attraction. In solid form, ionic compounds form a crystal lattice structure, contributing to their high melting and boiling points. NaA, as an ionic compound mentioned in our exercise, dissociates in water into sodium ions (Na⁺) and the anion (A⁻) of a weak acid. This dissociation is crucial during titration, as it allows the interaction with the added HCl, enabling precise quantitative measurements. Ionic compounds are typically:

• Good conductors of electricity when dissolved in water, due to free-moving ions.
• Possessing distinct and robust structures.

Understanding the nature of ionic compounds helps predict their behavior in various chemical processes and reactions, such as dissolution and titration.