Question

A \(0.7026 \mathrm{g}\) sample of an unknown acid requires \(40.96 \mathrm{mL}\) of \(0.1158 \mathrm{M} \mathrm{NaOH}\) for neutralization to a phenolphthalein end point. There are \(1.22 \mathrm{mL}\) of \(0.1036 \mathrm{M}\) HCl used for back-titration. a. How many moles of OH \(^{-}\) are used? How many moles of \(\mathrm{H}^{+}\) from HCl? ____ moles OH \(^{-}\) ______ moles \(\mathrm{H}^{+}\) b. How many moles of \(\mathrm{H}^{+}\) are there in the solid acid? (Use Eq. 5.) ______ moles \(\mathrm{H}^{+}\) in solid c. What is the molar mass of the unknown acid? (Use Eq. 4.) ______ \(g\)

Short answer

a. \(0.00474 \, \text{moles}\) OH\(^{-}\), \(0.000126 \, \text{moles}\) H\(^{+}\) from HCl b. \(0.00462 \, \text{moles}\) H\(^{+}\) in solid acid c. Molar mass of the unknown acid is \(152.03 \, \text{g/mol}\).

Step-by-step solution

Calculate the moles of OH-

The provided volume of NaOH is 40.96 mL and concentration is 0.1158 M. The number of moles of OH- can be found using the formula M = n/V, which can be rearranged to find n as n = M × V. So, \(n(OH^-) = 0.1158 \, \text{M} \times 0.04096 \, \text{L} = 0.00474 \, \text{moles}\).

Calculate the moles of H+ from HCl

The provided volume of HCl is 1.22 mL and concentration is 0.1036 M. Similarly, we can find n(HCl) as \(n(HCl) = 0.1036 \, \text{M} \times 0.00122 \, \text{L} = 0.000126 \, \text{moles}\).

Calculate the moles of H+ in the solid acid

According to Equation 5, the number of H+ ions in the solid acid is given by the difference between the moles of OH- and the moles of HCl. So, we can find n(H+) in solid as by plug our findings from step 1 and 2 into Eq. 5: \(n(H^+) = 0.00474 - 0.000126 = 0.00462 \, \text{moles}\).

Calculate the molar mass of the unknown acid

According to Equation 4, the molar mass of a substance is the ratio of mass to the number of moles. Given the mass of the solid acid is 0.7026 g and from step 3 we have the moles, we plug these values to find the molar mass using Eq.4: \(\text{Molar mass} = 0.7026 \, \text{g} / 0.00462 \, \text{moles} = 152.03 \, \text{g} / \text{mol}\).\ This is the molar mass of the unknown acid.

Key concepts

Phenolphthalein Endpoint

In acid-base titrations, the **phenolphthalein endpoint** is a crucial concept. Phenolphthalein serves as an indicator to signify the end of a titration by changing color, typically from colorless to pink in a basic solution. This occurs when the pH of the solution reaches about 8.2 - 10, indicating that all acidic protons have reacted with the base.
This endpoint usually appears slightly beyond the point of complete neutralization due to the behavior of the indicator itself, meaning it's after the exact equivalence point. To improve accuracy:

• Always add the titrant slowly as you approach the endpoint.
• Listen for subtle changes; the shift to pink should appear gradually with the slightest addition of base.

Understanding phenolphthalein's role helps prevent overtitrating and ensures a proper calculation of the unknown's concentration.

Molar Mass Calculation

The **molar mass calculation** of an unknown acid is a fundamental task in chemistry, allowing us to relate mass to moles. Molar mass is calculated by dividing the mass of the compound by the number of moles it contains. In this exercise, the mass of the acid is given as 0.7026 grams.
After identifying the number of moles of H⁺ ions, which in this case is 0.00462 moles, you calculate the molar mass using the formula:\[\text{Molar Mass} = \frac{\text{mass}}{\text{moles}}\]
Plugging in the values yields:\[= \frac{0.7026 \, \text{g}}{0.00462 \, \text{moles}} = 152.03 \, \text{g/mol}\]
This result implies that each mole of the acid weighs 152.03 grams, characterizing the acid's identity or class for further studies.

Neutralization Reaction

A **neutralization reaction** is a chemical reaction in which an acid and a base react to form water and a salt, effectively "neutralizing" each other.
In this titration problem, this concept is visible when NaOH (a base) reacts with an unknown acid. As the NaOH neutralizes the acid, it consumes the hydrogen ions (H⁺), producing water. The general equation for such a process is:\[\text{Acid (HA) + Base (MOH)} \rightarrow \text{Water} + \text{Salt (MA)}\]To find the leftover H⁺ in the solid, after using NaOH, a back titration with HCl was performed. The number of moles of OH⁻ used is initially calculated, and the remaining H⁺ moles adjusted using the moles of H⁺ from HCl, confirming the thoroughness of the neutralization process. This clear understanding of neutralization aids in accurate experimental outcomes and reinforces stoichiometric calculation skills.