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 molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) 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.

Short answer

(a) The molar mass is 94.48 g/mol. (b) The \( K_a \) of the acid is \( 1.29 \times 10^{-5} \). (c) The acid might be butanoic acid.

Step-by-step solution

Calculate Moles of NaOH

Use the formula \( n = C \times V \) to calculate the moles of \( \mathrm{NaOH} \). Here, \( C = 0.0500 \, \mathrm{M} \) and \( V = 22.10 \, \mathrm{mL} = 0.02210 \, \mathrm{L} \). Thus, the moles of \( \mathrm{NaOH} \) are:\[n = 0.0500 \, \mathrm{M} \times 0.02210 \, \mathrm{L} = 0.001105 \, \text{moles}\]Since the acid is monoprotic, the moles of acid are equal to the moles of \( \mathrm{NaOH} \) used at the endpoint.

Calculate Molar Mass of the Acid

Molar mass is calculated using \( \text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} \). The mass of acid is \( 0.1044 \, \text{g} \) and the moles, as calculated are \( 0.001105 \, \text{moles} \). So, the molar mass is:\[\text{Molar Mass} = \frac{0.1044 \, \text{g}}{0.001105 \, \text{moles}} = 94.48 \, \text{g/mol}\]

Calculate Halfway Moles

When 11.05 mL of base has been added, it is halfway to the equivalence point because 11.05 mL is half of 22.10 mL. At halfway, \( [\text{HA}] = [\text{A}^-] \), so \( \text{pH} = \text{pK}_a \).

Determine pKa and Ka

Since the \( \mathrm{pH} \) at halfway is \( 4.89 \), it follows that \( \text{pK}_a = 4.89 \). Convert \( \text{pK}_a \) to \( K_a \) using the formula:\[K_a = 10^{-\text{pK}_a} = 10^{-4.89} = 1.29 \times 10^{-5}\]

Suggest Identity of the Acid

Based on the calculated molar mass (94.48 g/mol) and \( K_a \) value \((1.29 \times 10^{-5})\), search Appendix D for acids with similar characteristics. An acid such as butanoic acid, which has a similar molar mass and \( K_a \) value, could be the identity.

Key concepts

Molar Mass Calculation

To find the molar mass of a substance, you need the mass of the sample and the number of moles it contains. In the context of acid-base titration, when you titrate a monoprotic acid using a base like NaOH, you can calculate the number of moles of acid neutralized by the base. This is because a monoprotic acid releases one hydrogen ion per molecule, reacting on a one-to-one basis with the base. The formula for molar mass is:\[\text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}}\]In the original exercise, the mass of the unknown acid is given as 0.1044 g. By determining the moles of NaOH used at the endpoint of the titration, which in turn equals the moles of the unknown acid, you can use the formula to find the molar mass. Here, 0.001105 moles of NaOH react completely with the acid to result in a molar mass of 94.48 g/mol.

pH and pKa Relationship

Understanding the relationship between pH and pKa is crucial in acid-base chemistry. The pH of a solution indicates its acidity or basicity, while pKa is a specific value for an acid describing its strength.The Henderson-Hasselbalch equation relates these two as follows:\[\text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right)\]When the concentration of the acid \(\text{HA}\) equals the concentration of its conjugate base \(\text{A}^-\), the term \(\log\left(\frac{[A^-]}{[HA]}\right)\) becomes zero, meaning \(\text{pH} = \text{pKa}\). This usually happens at the halfway point to the equivalence point in a titration.In the exercise, after adding 11.05 mL of NaOH, the pH was measured to be 4.89. This implies that at this halfway point, the pKa is equal to 4.89.

Equivalence Point

The equivalence point in a titration is reached when the number of moles of acid equals the number of moles of base added. It's an important milestone because it indicates that the acid has been completely neutralized by the base. In the exercise, the equivalence point occurs at 22.10 mL of NaOH. At this stage, both the acid and base are in stoichiometric balance, and no excess of either is present. The \(\text{pH}\) at this point is not directly given, but knowing that the molar ratio is 1:1 helps confirm the equivalence of moles used, which was calculated in the steps.

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton (hydrogen ion) per molecule in an aqueous solution. Examples include hydrochloric acid (\( \text{HCl} \) ) and acetic acid (\( \text{CH}_3\text{COOH} \) ). Monoprotic acids are simpler to titrate compared to polyprotic acids, which can donate more than one proton.In the original problem, a monoprotic acid is being titrated. This simplifies calculations, as it reacts on a one-to-one basis with the base used, typically a strong base like NaOH. The direct 1:1 stoichiometric reaction allows for straightforward calculations of the amount of acid present based on the volume and molarity of the titrant used.This property also aids in identifying the acid when combined with other data such as molar mass and \(K_a\) , as seen in the attempt to determine the identity of the unknown acid, potentially as butanoic acid, based on these calculations.