Advanced Study Assignment 1. 7.2 mL of 6.0MNaOH are diluted with water to a volume of 400.0 mL. You are asked to find the molarity of the resulting solution. a. How many moles of NaOH are there in 7.2 mL of this solution? moles b. Determine the molarity of the diluted NaOH solation. 2. In an acid-base titration, 21.16 mL of an NaOH solution are needed to neutralize 20.04 mL of a 0.0997M HCl solution. To find the molarity of the NaOH solution, we can ese the following procedure: a. Find the value of M_9 ∗ , for the HCl solution. b. Find the value of MOg_ay – for the NaOH solution. c. Obtain M_fyour from M^- _OH^− Experiment w24: Rase Standardizokion/Molar Mass of an Ae 3. A 0.3012 g sample of an unknown menoprotic acid requires 24.13 mL of a 0.0944M NaOH solution दor neutralization to a phenolphthaleln end point. There are 0.32mi, of 0.0997M/NC issed for back titration. 3. How many moles of oH are used? How many moles of H − from HCF? aholes DH moles H^* b. How many mioles of H^- are there in the solid acid? (Use Bq:.5) moles tut c. What is the molar mass of the unknown acid? (Use Eq 4)
The Correct Answer and Explanation is :
Let’s break down the assignment step by step to find the required values and provide explanations.
Problem 1: Dilution of NaOH Solution
1a. Moles of NaOH in 7.2 mL of 6.0 M NaOH:
To calculate the number of moles of NaOH in the solution, we use the formula:
[
\text{Moles} = \text{Molarity} \times \text{Volume (L)}
]
Convert 7.2 mL to liters:
[
7.2 \, \text{mL} = 0.0072 \, \text{L}
]
Now, plug in the values:
[
\text{Moles of NaOH} = 6.0 \, \text{M} \times 0.0072 \, \text{L} = 0.0432 \, \text{moles}
]
1b. Molarity of the diluted NaOH solution:
After dilution, the total volume is 400.0 mL, or 0.400 L. The molarity of the diluted solution can be found using the formula:
[
\text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}}
]
Substituting the values:
[
\text{Molarity of diluted NaOH} = \frac{0.0432 \, \text{moles}}{0.400 \, \text{L}} = 0.108 \, \text{M}
]
Problem 2: Titration of HCl with NaOH
2a. Moles of HCl:
To find the moles of HCl, we use the formula:
[
\text{Moles} = \text{Molarity} \times \text{Volume (L)}
]
Convert 20.04 mL to liters:
[
20.04 \, \text{mL} = 0.02004 \, \text{L}
]
Now, plug in the values:
[
\text{Moles of HCl} = 0.0997 \, \text{M} \times 0.02004 \, \text{L} = 0.0020008 \, \text{moles}
]
2b. Moles of NaOH:
Since the reaction between NaOH and HCl is a 1:1 mole ratio, the moles of NaOH required to neutralize the HCl is also 0.0020008 moles.
2c. Molarity of NaOH:
The molarity of the NaOH solution can be found using its volume (21.16 mL or 0.02116 L):
[
\text{Molarity of NaOH} = \frac{0.0020008 \, \text{moles}}{0.02116 \, \text{L}} \approx 0.0945 \, \text{M}
]
Problem 3: Back Titration of Unknown Monoprotic Acid
3a. Moles of NaOH used:
The total moles of NaOH initially used in the reaction is:
[
\text{Moles of NaOH} = 0.0944 \, \text{M} \times 0.02413 \, \text{L} = 0.002275 \, \text{moles}
]
3b. Moles of NaOH used in back titration:
[
\text{Moles of NaOH (back titration)} = 0.0997 \, \text{M} \times 0.00032 \, \text{L} = 0.000031904 \, \text{moles}
]
3c. Moles of acid:
To find the moles of the monoprotic acid, we can subtract the moles of NaOH used in back titration from the total moles used initially:
[
\text{Moles of Acid} = \text{Moles of NaOH} – \text{Moles of NaOH (back titration)} = 0.002275 – 0.000031904 \approx 0.002243 \, \text{moles}
]
3d. Molar mass of the unknown acid:
The molar mass can be calculated using the mass of the acid sample:
[
\text{Molar Mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.3012 \, \text{g}}{0.002243 \, \text{moles}} \approx 134.00 \, \text{g/mol}
]
Summary of Answers
- Moles of NaOH: 0.0432 moles
- Molarity of diluted NaOH: 0.108 M
- Moles of HCl: 0.0020008 moles
- Molarity of NaOH: 0.0945 M
- Moles of unknown acid: 0.002243 moles
- Molar mass of the unknown acid: 134.00 g/mol
Explanation
In this assignment, we systematically used stoichiometric relationships and dilution principles to find concentrations and moles of substances involved in acid-base reactions. Molarity calculations were performed by converting volumes from milliliters to liters and applying the fundamental formula of concentration. The back titration provided a means to determine the amount of acid based on neutralization reactions, leveraging both the initial and back titration data. Understanding these concepts is crucial in analytical chemistry, especially for quantifying unknown substances and determining their properties.