Question

A \(10.00-g\) 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 mL of solution and was then titrated with 0.100 \(M\) HCl. After 500.0 mL HCl was added, the pH was \(5.00 .\) The experimenter found that 1.00 L of \(0.100 M\) HCl was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(p\) H of the solution at the stoichiometric point of the titration.

Short answer

a. To find the molar mass of NaA, remember that in a titration with a strong acid, the moles of base (NaA in this case) initially present would be equal to the moles of acid (HCl) required to reach the equivalence point. This means that 1.00 L of 0.100 M HCl was needed to neutralize the sample of NaA, so there were 0.1 moles of NaA in the original sample. Since this 0.1 moles weighed 10.00 g, the molar mass must be \( \frac{10.00 g}{0.1 mol} = 100.00 g/mol \). b. To find the pH at the stoichiometric point, we must consider what is present in solution at this point- namely, the anion A^- in solution. The reaction occurring will be \( A^- + H_2O <-> HA + OH^- \) and the pH of the solution will depend on the equilibrium that results from this reaction. Given that the pH of the solution was 5.00 after 500 ml of 0.100 M HCl had been added and that the volume of the solution was now 600 ml, we can first calculate the concentrations of all the species in solutions and by taking the negative logarithm of these concentrations, we can find the pH. The reaction proceeds to give [H+] = 10^-4.83 M and hence, pH = 4.83.

Step-by-step solution

Find moles of HCl and NaA

First, we'll find the moles of HCl added at the stoichiometric point and moles of NaA in the original solution. Given: Volume of HCl required to reach the stoichiometric point = 1.00 L Concentration of HCl = 0.100 M Moles of HCl = Concentration × Volume Moles of HCl = 0.100 moles/L × 1.00 L Recall that for a complete reaction, moles of NaA = moles of HCl.

Calculate the molar mass of NaA

Now that we have the moles of NaA, we can find its molar mass. Given: Mass of NaA = 10.00 g Moles of NaA = Moles of HCl = 0.100 moles Molar mass of NaA = (Mass of NaA)/(Moles of NaA) = (10.00 g)/(0.100 moles) = 100.00 g/mol Answer to part (a): The molar mass of NaA is 100.00 g/mol.

Calculate the pH at the stoichiometric point

To calculate the pH at the stoichiometric point, we need to find the concentration of the A^(-) anion. The stoichiometric point reached means that all NaA has reacted with HCl, so the remaining solution will have A^(-): \(NaA + HCl \rightarrow NaCl + HA\) At this point, we can use the given pH value before the stoichiometric point was reached to find the Ka (Acid dissociation constant) of the weak acid HA. We must remember that the concentration of A^(-) is equal to the concentration of H+ at this point. Given: pH_after_500mL = 5.00 pKa = -log(Ka) From the given pH value, we get: [H+] = 10^(-5.00) = 1.0 × 10^(-5) M Following the formula, pKa = -log([H+] × [A^(-)]/[HA]) Adding 500 mL of HCl to 100 mL of NaA solution, the total volume will be 600 mL or 0.6 L. We calculate the concentration of A^(-), which is equal to [H+] at this point. [A^(-)] = (0.100 M × 0.500 L) / 0.6 L = 0.0833 M We can find the concentration of HA by considering that in the reaction, 1 mole of NaA is converted to 1 mole of HA. [HA] = (0.100 M × 0.100 L) / 0.6 L = 0.0167 M Now we can calculate pKa: 5.00 - pH_after_500mL = -log((1.0 × 10^(-5) × 0.0833)/0.0167) pKa = 4.83 Therefore, Ka = 10^(-4.83) At the stoichiometric point, the remaining solution composition is [A^(-)] = [HA] = 0.100 M (since they are produced in equimolar amounts, and NaCl does not affect the pH). We can now use the Ka expression to find [H+] at the stoichiometric point: Ka = [H+][A^(-)]/[HA] [H+] = Ka * [HA]/[A^(-)] = 10^(-4.83) * 0.100/0.100 = 10^(-4.83) M Now, we can find the pH at the stoichiometric point by taking the negative logarithm of the [H+] concentration: pH = -log10([H+]) = -log10(10^(-4.83)) = 4.83 Answer to part (b): The pH at the stoichiometric point of the titration is 4.83.

Key concepts

Ionic Compound

Ionic compounds are formed when positively charged ions, or cations, are attracted to negatively charged ions, or anions. This forms a chemical bond, creating a compound that is usually quite stable. In the case of our compound, NaA, sodium (Na) is the cation, and A\(^-\), which represents a generic anion, is the negatively charged counterpart. Together, they form NaA, which dissolves in water to dissociate into its individual ions. This means in solution, Na\(^+\) and A\(^-\) are free to interact with other substances, such as HCl in our titration process.

Molar Mass

Molar mass is a fundamental property of a substance, expressing the mass of one mole of that substance in grams per mole (g/mol). It allows chemists to count particles by weighing them. In this exercise, we're given a 10.00-g sample of NaA. By knowing that 0.100 moles of NaA were used in the reaction, we can calculate its molar mass with the formula:
\[ \text{Molar mass} = \frac{\text{mass of NaA}}{\text{moles of NaA}} = \frac{10.00 \text{ g}}{0.100 \text{ moles}} = 100.00 \text{ g/mol} \]
Thus, the molar mass of NaA is 100.00 g/mol, which helps us understand how much of the compound we have when given a particular mass, allowing for accurate measurements in chemical reactions.

Stoichiometric Point

The stoichiometric point in a titration is where the amount of titrant added is exactly enough to completely react with the substance being titrated. In this problem, the stoichiometric point is reached when 1.00 L of 0.100 M HCl has been added. This means all NaA has reacted with HCl, converting entirely to NaCl and HA:
\[ \text{NaA + HCl} \rightarrow \text{NaCl + HA} \]
At this point, the volumes and concentrations tell us the proportions of the reactants and products present in the solution. The stoichiometric point is crucial in determining other properties like the pH of the solution, as the concentration of ions changes.

Acid Dissociation Constant

The acid dissociation constant, denoted as \(K_a\), is a vital value that describes the strength of a weak acid in solution. It tells us how well an acid donates protons (H\(^+\) ions) in a reaction. For our weak acid, HA, generated when NaA reacts with HCl, we want to understand its dissociation:

• The pH after adding 500.0 mL of HCl was 5.00, giving us [H\(^+\)] = \(10^{-5.00}\) M.
• Using \(pK_a = -\log K_a\), we calculated \(K_a\) based on proton and anion concentrations derived from solution volumes.

This calculation is helpful because understanding \(K_a\) lets us predict the behavior of acids in various chemical settings, and is essential for calculating pH at different phases of the titration.