/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 125 A solution is prepared by dissol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solution is prepared by dissolving 0.6706 g oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make 100.0 \(\mathrm{mL}\) of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the oxalic acid solution?

Short Answer

Expert verified
The final molarity of the oxalic acid solution can be calculated using the following steps: 1. Calculate moles of oxalic acid in the initial 100 mL solution: \(Moles = \frac{0.6706\ g}{90.04\ g/mol} \approx 0.00745\ mol\). 2. Determine the concentration of oxalic acid in the initial 100 mL solution: \(Molarity = \frac{0.00745\ mol}{0.1\ L} = 0.0745\ M\). 3. Calculate moles of oxalic acid in the 10 mL aliquot: \(Moles\ in\ aliquot = 0.0745\ M \times 0.01\ L = 0.000745\ mol\). 4. Determine the final molarity of oxalic acid in the final 250 mL diluted solution: \(Final\ Molarity = \frac{0.000745\ mol}{0.25\ L} = 0.00298\ M\). Thus, the final molarity of the oxalic acid solution is approximately \(0.00298\ M\).

Step by step solution

01

1. Calculate the moles of oxalic acid in the initial 100 mL solution.

To determine the moles of oxalic acid, we'll use the molecular weight of oxalic acid, and the given mass: Molecular weight of oxalic acid, H2C2O4 is \(2 \times 1.01 (H) + 2 \times 12.01 (C) + 4 \times 16.00 (O) = 90.04\ g/mol\). Given mass of oxalic acid is 0.6706 g. Now, use the formula: Moles = \(\frac{Mass}{Molecular\ weight}\) Moles = \(\frac{0.6706}{90.04}\)
02

2. Determine the concentration of oxalic acid in the initial 100 mL solution.

Since we now have the moles of oxalic acid, we can find the initial concentration using the volume of the initial 100 mL solution: Molarity = \(\frac{Moles}{Volume(in\ L)}\) Molarity = \(\frac{Moles}{0.1\ L}\)
03

3. Calculate the moles of oxalic acid in the 10 mL aliquot.

The aliquot is a portion of the original solution, with the same concentration. To find the moles of oxalic acid in the aliquot, multiply the concentration from step 2 by the aliquot volume in liters: Moles in aliquot = (Molarity in initial solution) \(\times\) (Volume of aliquot)
04

4. Determine the final molarity of oxalic acid in the final 250 mL diluted solution.

Now, we'll find the final molarity by dividing the moles of oxalic acid in the aliquot by the final volume of the diluted solution (250 mL): Final Molarity = \(\frac{Moles\ in\ aliquot}{Final\ volume(in\ L)}\) Final Molarity = \(\frac{Moles\ in\ aliquot}{0.25\ L}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Dilution
Solution dilution is a common laboratory process used to reduce the concentration of a solute in a solution. In simple terms, it involves adding more solvent to a concentrated solution to make it less concentrated. This is particularly useful when the desired concentration for a task or experiment is lower than that of the prepared solution.

Consider an example where you want to dilute an oxalic acid solution. If you start with a 10 mL solution and dilute it to a total volume of 250 mL, you are diluting the solution by a factor of 25. The concentration of the acid in the final solution is then much less than it was originally.
  • Start with the initial concentration of the solution.
  • Determine the volume to which you want to dilute.
  • Apply the dilution formula: \( C_1V_1 = C_2V_2 \)
Remember, the number of moles of solute remains constant before and after dilution; it's just the concentration that changes.
Oxalic Acid
Oxalic acid, represented chemically as \( \text{H}_2\text{C}_2\text{O}_4 \), is a simple dicarboxylic acid that occurs naturally in many plants. It's commonly used in chemical laboratories.

This acid is often in the form of a white crystalline solid and exhibits a molecular weight of 90.04 g/mol. Due to its acidic nature, it can act as a bleaching agent and is often used in cleaning and restoration processes. In laboratory settings, oxalic acid solutions are used for titrations and other analytical purposes. Its well-known properties make it an ideal choice for experiments involving acid-base reactions.
  • Typically found as a dihydrate (two water molecules attached).
  • Moderately soluble in water.
  • Used in rust removal due to its capability to form stable, complex anions with metal ions.
When handling oxalic acid, safety precautions are vital due to its potential toxicity if ingested or improperly handled.
Moles Calculation
The concept of moles is central to chemistry, serving as a bridge between atom-scale measurements and larger, laboratory-scale experiments. Calculating moles starts with the given mass and the known molecular weight of the compound.

In our case of oxalic acid, for example, you calculate the moles by dividing the sample's weight by its molecular weight using the formula:
\[\text{Moles} = \frac{\text{Mass}}{\text{Molecular Weight}}\]
For oxalic acid, this calculation involves dividing 0.6706 grams by 90.04 g/mol. The result represents the number of moles of oxalic acid present in your solution.
  • Always start by identifying the molecular weight based on atomic masses.
  • Use precise mass measurements for accurate calculations.
  • Remember, the mole is a fundamental unit in chemistry, providing a method to count particles by weighing them.
Calculating moles is essential for finding concentrations and understanding chemical reactions.
Concentration Determination
Determining the concentration of a solution is vital for many chemical processes, as it tells you how much solute is present in a certain amount of solvent. Molarity (\( M\)) is the most common way to express concentration and is defined as moles of solute per liter of solution.

In the exercise example, the concentration of oxalic acid is initially calculated for a 100 mL solution, which is then used to determine the concentration after dilution. The process involves calculating and understanding molarity:
\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]
For the final diluted solution, find the moles remaining in the aliquot and divide by the new total volume in liters to get the final molarity.
  • Concentration helps in understanding solution properties.
  • Makes it easier to compare different solutions quantitatively.
  • Essential for stoichiometric calculations and reactions.
Concentration tells us about the solution's "strength" and its utility in different chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?

You have a sugar solution (solution A) with concentration x. You pour one- fourth of this solution into a beaker, and add an equivalent volume of water (solution B). a. What is the ratio of sugar in solutions A and B? b. Compare the volumes of solutions A and B. c. What is the ratio of the concentrations of sugar in solutions A and B?

Consider the reaction of 19.0 g of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and 29.0 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\) . A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Many plants are poisonous because their stems and leaves contain oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) or sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) . When ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, \(\mathrm{CaCl}_{2},\) in aqueous solution.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.