Question

A \(10.00-\mathrm{g}\) sample of the ionic compound 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 \(0.100 M\) 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 \(\mathrm{NaA}\) ? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.

Short answer

The molar mass of NaA is 100.00 g/mol. The pH of the solution at the stoichiometric point is 2.99.

Step-by-step solution

Calculate moles of HCl at stoichiometric point

\ To find the molar mass of NaA, we first need to determine the moles of HCl at the stoichiometric point. We are given the volume (1.00L) and concentration (0.100M) of HCl at the stoichiometric point: Moles of HCl = volume × concentration Moles of HCl = 1.00 L × 0.100 M = 0.100 mol

Determine moles of A¯ in the solution at the stoichiometric point

\ At the stoichiometric point, the number of moles of A¯ will be equal to the number of moles of H⁺ from HCl. So, the moles of A¯ in the solution are 0.100 mol.

Calculate the concentration of A¯ in the solution before titration

\ The 10.00g sample of the NaA compound was dissolved in 100.0 mL of water before performing the titration. At this point, the concentration of A¯ is equal to the moles of A¯ divided by the volume of the solution in liters: Concentration of A¯ = moles of A¯ / volume of solution Concentration of A¯ = 0.100 mol / 0.100 L = 1.00 M

Find the molar mass of NaA

\ We now have the concentration of A¯ in the solution and the mass of the compound. We can use these values to calculate the molar mass of NaA: Molar mass of NaA = mass of NaA / moles of A¯ Molar mass of NaA = 10.00 g / 0.100 mol = 100.00 g/mol The answer for part a is: The molar mass of NaA is 100.00 g/mol.

Determine the concentration of HA at the stoichiometric point

\ At the stoichiometric point, the moles of A¯ and HA are equal. Since we already know the moles of A¯ at the stoichiometric point (0.100 mol), we can use the volume of the solution at that point to calculate the concentration of HA: Concentration of HA = moles of HA / volume of solution at stoichiometric point Concentration of HA = 0.100 mol / (0.100 L + 1.00 L) = 0.100 mol/1.10 L = 0.0909 M

Calculate the pH at the stoichiometric point

\ At the stoichiometric point, the solution contains only the weak acid HA. We can use its concentration (0.0909 M) and the given pH (5.00) at a certain point of the titration (after addition of 500.0 mL of HCl) to determine the Ka of the weak acid: \[Ka = 10^{-pKa}\] \[pKa = pH - \log{(\frac{[HA]}{[A^-]})}\] \[pKa = 5.00 - \log{(\frac{0.0909 M}{0.100 M})}\] \[pKa \approx 4.95\] \[Ka \approx 1.12 \times 10^{-5}\] Now we can use the Ka value and the concentration of HA to calculate the H⁺ concentration at the stoichiometric point: \[Ka = \frac{[H^+][A^-]}{[HA]}\] \[1.12 \times 10^{-5} = \frac{[H^+]^2}{0.0909 M}\] \[H^+ = \sqrt{1.12 \times 10^{-5} \times 0.0909 M}\] \[H^+ \approx 1.02 \times 10^{-3}\] Finally, we can calculate the pH at the stoichiometric point using the H⁺ concentration: \[pH = -\log{[H^+]}\] \[pH \approx -\log{(1.02 \times 10^{-3})}\] \[pH \approx 2.99\] The answer for part b is: The pH of the solution at the stoichiometric point is 2.99.

Key concepts

Ionic Compound

An ionic compound is made up of ions, which are atoms or molecules that have gained or lost one or more electrons, resulting in a net charge. In the context of this exercise, NaA is the ionic compound being discussed.
Here, Na represents sodium, which is a metal, and \(A^-\) represents the anion from a weak acid. This anion carries a negative charge. Sodium, a metal, usually forms a cation by losing one electron, resulting in \(Na^+\).
The ionic bond arises due to the attraction between Na^+ and A^-. These types of compounds dissolve in water, dissociating into their individual ions, making them critical for titration exercises.
Titration balances these dissociated ions, allowing us to calculate important properties like molar mass and pH.

Stoichiometric Point

In a titration, the stoichiometric point—also referred to as the equivalence point—is the point at which the amount of titrant added is exactly enough to neutralize the analyte solution. Here, the stoichiometric point is reached when the amount of HCl added corresponds precisely to the amount of the conjugate base (A\(^-\)) originally present in solution.
Reaching this point implies that all the initial ions have reacted, which lets us derive the number of moles of the initial compound based on the reaction's equation:

• Number of moles of H\(^+\) (from HCl) matches the moles of A\(^-\).
• Allows calculation of the initial concentration of species in the solution.

This is crucial for determining the molar mass of compounds and the properties of the solution after reaction.

Weak Acid

A weak acid only partially ionizes in solution, which means it does not completely dissociate into its ions. In this exercise, the weak acid in question is represented as HA.
The dissociation in water can be represented as follows: HA \( \rightleftharpoons \) H\(^+\) + A\(^-\).
With weak acids, the equilibrium is dynamic and exists between undissociated molecules and free ions. This incomplete ionization causes the weak acid to have a higher pH than a strong acid of similar concentration.
The dissociation constant (Ka) plays a pivotal role, often utilized in calculations of pH within titration problems.Understanding weak acids is essential in solving the pH changes during titration, as observed when titrating NaA with HCl.

pH Calculation

The pH of a solution is a measure of its hydrogen ion concentration and is calculated using pH = -log[H^+]. During titration, pH changes and leads to vital information about the solution composition. In our problem, once titration was at the stoichiometric point, HA's presence impacted the pH.
The process for calculating pH in this scenario involves using the acid dissociation constant (Ka):

• Set up equilibrium with [H^+] = [A^-], knowing [HA] from titration.
• Utilize Ka to find [H^+] concentration, assuming equilibrium conditions.

Finally, translate the [H^+] into pH to find our solution's acidity level at any point. This highlights the importance of equilibrium constants in correlating concentration of ions to pH change.