Question

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Short answer

The molecular weight of the unknown monoprotic acid is approximately 94.48 g/mol, and its \(K_{a}\) value is approximately 1.68 × 10^-6. Comparing these results with the values from Appendix D, the acid might be benzoic acid (Molecular weight: 122.12 g/mol, \(K_{a} = 6.46 × 10^{-5}\)). Although the values are not an exact match, they are reasonably close, suggesting benzoic acid as a plausible identity for the unknown acid.

Step-by-step solution

(a) Calculate moles of NaOH

First, let's calculate the moles of NaOH used to neutralize the acid. We are given the volume of NaOH (22.10 mL) and the concentration (0.0500 M). We can use this information to find the moles of NaOH: moles of NaOH = volume × concentration moles of NaOH = 22.10 mL × 0.0500 M moles of NaOH = 1.105 × 10^-3 mol

(a) Calculate moles of the acid

Given that the acid is monoprotic, 1 mole of NaOH will react with 1 mole of the acid in a 1:1 ratio. So, the moles of the acid are equal to that of moles of NaOH: moles of the acid = 1.105 × 10^-3 mol

(a) Calculate the molecular weight of the acid

We are given the mass of the acid sample (0.1044 g). We can now use the moles of the acid to calculate its molecular weight: molecular weight = mass / moles molecular weight = 0.1044 g / (1.105 × 10^-3 mol) molecular weight ≈ 94.48 g/mol

(b) Calculate the concentration of H+

Now let's calculate the H+ concentration in the solution after the addition of 11.05 mL of the base (halfway to the equivalence point). We can use the pH value given (4.89) to find the concentration of H+: [H+] = 10^(-pH) [H+] ≈ 1.29 × 10^-5 M

(b) Calculate the original concentration of the acid

As we are at the halfway point to the equivalence, the concentration of the undissociated acid, HA, is equal to the initial concentration of the acid minus the concentration of H+ ions. Initial concentration of acid = moles of acid / initial volume of the solution Initial concentration of the acid ≈ (1.105 × 10^-3) / 0.01105 L ≈ 0.100 M At the halfway point, the concentration of the HA and A- (conjugate base) are equal. Therefore, [HA] = [A-]. Let x be the concentration of the dissociated acid (the loss in [HA] and equals to [H+]): [HA] = 0.100 - x [A-] = x At the halfway point, x = [H+] x = 1.29 × 10^-5 M

(b) Calculate the value of Ka

Now, we can set up the equilibria expression for the reaction between the monoprotic acid, HA, and water: Ka = ([H+][A-]) / [HA] As [HA] = [A-], we can substitute the values into the equation: Ka = (1.29 × 10^-5)^2 / (0.100 - 1.29 × 10^-5) ≈ 1.68 × 10^-6

(c) Identify the unknown acid

Now, we have the molecular weight and Ka for the acid. We can compare these values to those in Appendix D to suggest the identity of the acid. Based upon the molecular weight (≈ 94.48 g/mol) and Ka value (≈ 1.68 × 10^-6), the acid might be: Benzoic acid (Molecular weight: 122.12 g/mol, Ka = 6.46 × 10^-5) Although the values are not an exact match, they are reasonably close. The choice of benzoic acid seems plausible. It is important to note that a more precise identification would require additional experimental data, such as IR or NMR spectra.

Key concepts

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton (hydrogen ion) per molecule when it dissolves in water. This simpler behavior compared to polyprotic acids, which can donate multiple protons, makes titration calculations involving monoprotic acids relatively straightforward.
In titration, knowing that an acid is monoprotic is essential because it informs the stoichiometry of the neutralization reaction. For a monoprotic acid, the reaction with a base like NaOH occurs in a 1:1 ratio. This means one mole of the acid reacts with one mole of NaOH. This relationship is pivotal for determining the amount of acid or base needed to reach the equivalence point in a titration.

Molecular Weight Determination

The molecular weight of a compound is the sum of the atomic masses of its elements, measured in g/mol. Determining the molecular weight of an unknown acid using titration involves a few critical steps. First, calculate the moles of base (e.g., NaOH) used. Then, assuming a monoprotic acid, these moles are equal to the moles of acid because they react in a 1:1 ratio.
To find the molecular weight, you need the mass of the acid sample and the moles calculated from the titration. Use the equation: \[\text{Molecular Weight} = \frac{\text{Mass of Sample}}{\text{Moles of Acid}}\]Applying this to our example, if the mass of the acid is 0.1044 g and the moles of acid are calculated to be \(1.105 \times 10^{-3}\) mol, the molecular weight turns out to be approximately 94.48 g/mol. These calculations help in identifying the unknown acid by comparing the molecular weight with known values.

Acid Dissociation Constant (Ka)

The acid dissociation constant, \(K_a\), measures the strength of an acid in solution. It reflects the acid's ability to donate protons to bases. A higher \(K_a\) value indicates a stronger acid that dissociates more in solution. In titration, finding \(K_a\) involves calculating the concentration of hydrogen ions, \([H^+]\), usually determined from the pH: \[ [H^+] = 10^{-\text{pH}} \]For the given example, with a pH of 4.89 after halfway through the titration, \([H^+]\) is about \(1.29 \times 10^{-5}\) M.
Using the equilibrium expression \(K_a = \frac{[H^+][A^-]}{[HA]}\), you can substitute the values to find \(K_a\), knowing that at the halfway point, \([HA] = [A^-]\) and \([H^+]=[A^-]=x\). Thus, \(K_a\approx1.68 \times 10^{-6}\) using our example. This value is useful for identifying the unknown acid when compared to known acids.

Neutralization Reaction

A neutralization reaction involves the reaction of an acid with a base to produce water and a salt. In the context of titration, the base used might be a standard solution like NaOH, which interacts with the acid to find the equivalence point—the point where equal moles of acid and base have reacted.
The equation for a simple neutralization involving a monoprotic acid and NaOH can be shown as:\[\text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O}\]Understanding this essential reaction helps in performing precise titration calculations, as it indicates the endpoint when the pH changes dramatically. This reaction's stoichiometry also assures that the titration only involves equal molar amounts of acid and base, simplifying mole calculations for the titration's quantitative analysis.