Question

(a) \(\mathrm{A} .0 .1044\) -g sample of an unknown monoprotic acid requires 22.10 \(\mathrm{mL}\) of 0.0500 \(\mathrm{MNaOH}\) to reach the end point. What is the molar mass 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.

Short answer

In summary, the molar mass of the unknown monoprotic acid is 94.48 g/mol, and its Ka is \(1.29 \times 10^{-5}\). The identity of the acid does not match any common weak acids from the given table, but it might be close to propanoic or acetic acid, or a less common weak acid.

Step-by-step solution

Finding moles of NaOH and the unknown acid

To find the moles of the unknown acid, we first need to find the moles of NaOH used to reach the end point. We are given the volume (22.10 mL) and concentration (0.0500 M) of NaOH, which we can use to find the moles of NaOH. moles of NaOH = Molarity (NaOH) × Volume (NaOH) moles of NaOH = 0.0500 M × 0.02210 L = 0.001105 mol Since the unknown acid is monoprotic, the mole ratio between the acid and NaOH is 1:1. This means there are 0.001105 moles of the unknown acid present in the sample.

Determining the molar mass of the unknown acid

Now that we have the moles of the unknown acid, we can determine its molar mass using the given mass of the acid (0.1044 g): Molar mass (unknown acid) = mass (unknown acid) ÷ moles (unknown acid) Molar mass (unknown acid) = 0.1044 g ÷ 0.001105 mol = 94.48 g/mol

Finding the ion concentration at the given pH

Next, we need to find the ion concentration when the pH of the solution is 4.89 after adding 11.05 mL of the 0.0500 M NaOH solution. Use the pH formula: pH = -log[H+] Solve for [H+]: [H+] = 10^(-pH) [H+] = 10^(-4.89) = 1.29 × 10^(-5) M Now, we need to find the concentration of the acid ([A-]) in the solution. Since we know that at the halfway point of the titration, the concentrations of the acid and its conjugate base are equal. Therefore, [A-] = [H+].

Calculating the Ka of the unknown acid

Finally, we will calculate the Ka of the unknown acid using the given pH and ion concentrations. Ka = [H+][A-]/[HA] Since we are at the halfway point, [A-] = [H+] and [HA] = (moles of acid - moles of base) / total volume. In this case, it's (0.001105-0.0005525)/(0.01105+0.02210): [HA] = (0.001105-0.0005525)/(0.01105+0.02210) = 1.29 × 10^(-5) M Ka = (1.29 × 10^(-5))^2 / (1.29 × 10^(-5)) = 1.29 × 10^(-5)

Identifying the unknown acid

Now, using the Ka value of \(1.29 \times 10^{-5}\) and Appendix D or a table of Ka values for common weak acids, we can identify the unknown acid. The acid with a Ka value closest to ours is propanoic acid; acetic acid is another possible candidate (the two differ in their molecular weight). The molar masses of propanoic acid and acetic acid are around 74 g/mol and 60 g/mol. As the unknown acid was found to be 94.48 g/mol, the identity of the acid does not match any acid from the common weak acids list, but it could be close to propanoic or acetic acid, or it could be a less common weak acid not listed in the table.

Key concepts

Molar Mass Calculation

Understanding the concept of molar mass is fundamental to solving many chemistry problems, including the quantification of substances involved in chemical reactions. The molar mass is defined as the mass of one mole of a substance (usually in grams) and it is numerically equivalent to the substance's atomic or molecular weight.

For a monoprotic acid, where a single proton (H+) can be dissociated, the molar mass is particularly straightforward to calculate after an acid-base titration. In the example provided, the mass of the unknown monoprotic acid is given (0.1044 g), and through titration, we find the moles of acid that reacted with the base. To calculate the molar mass, we simply divide the mass of the acid sample by the number of moles of acid (moles are found by calculating moles of NaOH titrated, which equals to moles of the acid in a 1:1 reaction). The molar mass provides invaluable information about the substance's identity and purity, which is why it's crucial in analytical chemistry and synthesis.

Ka Calculation

When it comes to acid-base chemistry, the acid dissociation constant (Ka) is a pivotal quantity as it gives insight into the strength of an acid. Ka is a measure of how readily an acid donates a proton (H+) to the base in the solution.

In the titration example, the calculation of the Ka is performed at the halfway point of the titration. This is because, at this point, the concentrations of acid and conjugate base are equal, simplifying the calculation. We deduce the concentration of hydrogen ions ([H+]) from the pH, and since we're at the halfway point, [H+] is equal to the concentration of the conjugate base ([A-]). The Ka is then calculated by squaring the concentration of [H+] at this point. The understanding of Ka deepens our comprehension of weak acid behaviors and helps predict the outcomes of chemical reactions in which they take part.

Monoprotic Acid

Monoprotic acids are the simplest category among acids in that each molecule can donate only one proton to a base. This single-proton donation capability has a significant impact on the titration process and subsequent calculations. A classic example of a monoprotic acid is hydrochloric acid (HCl), where only one H+ ion is released into the solution.

In a titration setting, the knowledge that an acid is monoprotic simplifies the stoichiometry because it guarantees a 1:1 molar relationship with a monoprotic base, like NaOH. This piece of information is essential in our exercise because it ensures that the moles of NaOH used in the titration are equivalent to the moles of the unknown acid. Grasping the implications of a substance being monoprotic enables a more straightforward approach to various analytical techniques in chemistry, such as titrations, and facilitates a better understanding of acid-base reactions.