(a) What volume of a \(0.778 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (aq) solution should be diluted to \(150.0 \mathrm{~mL}\) with water to reduce its concentration to \(0.0234 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) ? (b) An experiment requires the use of \(60.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{NaOH}\) (aq). The stockroom assistant can only find a reagent bottle of \(2.5 \mathrm{M} \mathrm{NaOH}\) (aq). How is the \(0.50 \mathrm{M} \mathrm{NaOH}(\) aq \()\) solution to be prepared?

Short Answer

Expert verified
(a) The volume of the original Na2CO3 solution needed for dilution is 4.50 mL. (b) To prepare 60.0 mL of 0.50 M NaOH, 12.0 mL of the 2.5 M NaOH stock solution should be diluted.

Step by step solution

01

Understanding the Dilution Concept

Dilution involves adding water to a solution to lower its concentration without changing the amount of solute present. We use the dilution formula: \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume, while \(C_2\) and \(V_2\) are the concentration and volume of the diluted solution.
02

Calculate Volume Needed for Dilution of Na2CO3

To find the volume \(V_1\) of the original Na2CO3 solution needed to reach a new concentration \(C_2\) after dilution, rearrange the dilution formula to solve for \(V_1\): \(V_1 = \frac{C_2V_2}{C_1}\). Plug in \(C_1 = 0.778\text{ M}\), \(C_2 = 0.0234\text{ M}\), and \(V_2 = 150.0\text{ mL}\), then calculate \(V_1\).
03

Calculating the Exact Volume to Use

Calculate \(V_1 = \frac{0.0234 \times 150.0}{0.778}\) to get the volume \(V_1\) of the concentrated Na2CO3 solution required.
04

- Understanding the Preparation Concept

To prepare a diluted NaOH solution from a more concentrated stock, we also use the formula \(C_1V_1 = C_2V_2\). Here, we want to calculate the volume \(V_1\) of the 2.5 M NaOH solution that needs to be diluted to get 60.0 mL of a 0.50 M NaOH solution.
05

Calculate Volume Needed for Dilution of NaOH

Rearrange the dilution formula to solve for \(V_1\): \(V_1 = \frac{C_2V_2}{C_1}\), where \(C_1 = 2.5\text{ M}\), \(C_2 = 0.50\text{ M}\), and \(V_2 = 60.0\text{ mL}\). Calculate \(V_1\) to find out how much of the 2.5 M NaOH is needed.
06

Calculating the Exact Volume to Dilute

Calculate \(V_1 = \frac{0.50 \times 60.0}{2.5}\) to get the volume \(V_1\) of the 2.5 M NaOH solution required to make the 60.0 mL of 0.50 M NaOH solution.

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Key Concepts

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

Molarity
Molarity is a measure of the concentration of a solute in a solution and is expressed in moles per liter (M). It describes how many moles of a substance are present in one liter of a solution. For example, if you have a 1.0 M solution of sodium chloride, that means there is 1 mole of sodium chloride dissolved in every liter of solution. Molarity is a critical concept in chemistry because it allows for the quantification of reaction constituents, which is essential for predicting the outcomes of chemical reactions.

To calculate molarity, use the formula: \(M = \frac{n}{V}\), where \(M\) is molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters. This calculation is pivotal when mixing solutions to achieve a desired concentration for experimental protocols.
Dilution Formula
The dilution formula is essential for adjusting solution concentrations. It is represented as \(C_1V_1 = C_2V_2\), where \(C_1\) is the initial concentration, \(V_1\) the initial volume, \(C_2\) the final concentration, and \(V_2\) the final volume after dilution. This formula is based on the principle that the amount of solute in the solution remains constant before and after dilution. The formula enables the calculation of how much solvent (usually water) needs to be added to achieve a specific lower concentration or conversely, how much of a concentrated stock solution is required to prepare a more dilute solution. This tool is invaluable in the laboratory when preparing solutions for experiments where precise concentrations are key.
Concentration of Solutions
Concentration of solutions describes the amount of solute that is present in a given quantity of solvent or solution. It is a pivotal aspect of many chemical procedures and protocols as it affects reaction rates, yields, and can alter the nature of a reaction altogether. Concentration can be expressed in various ways such as molarity (M), mass percent, volume percent, and parts per million (ppm). Molarity, however, is the most commonly used unit in a laboratory setting for its ease of interconversion and direct use in stoichiometric calculations. It's important to understand how to manipulate concentrations to ensure the correct stoichiometry in reactions and to maintain safety standards when handling chemicals.
Volume Calculation
Volume calculation in the context of solution preparation is integral to the dilution process. To accurately prepare a desired solution, the volume of solvent to be added or the volume of stock solution to be used must be precisely calculated. The volume of a liquid can be measured in milliliters (mL) or liters (L), with 1,000 mL being equivalent to 1 L. When dealing with solutions, volumes need to be measured accurately using appropriate laboratory equipment such as graduated cylinders, pipettes, or burettes. Accurate volume calculation ensures that the correct molarity and concentration of the solution are maintained, which is critical for the success of chemical experiments.

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Most popular questions from this chapter

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

In a laboratory exercise, a student mixes \(2.5 .0 \mathrm{~g}\) of cthanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\), with \(150 \mathrm{~g}\) of water. (a) What is the mole fraction of ethanol in the solution? (b) What is the molality of cthanol in the solution?

(a) A chemist prepared a solution by dissolving \(1.230 \mathrm{~g}\) of \(\mathrm{KCl}\) in enough water to make \(150.0 \mathrm{~mL}\) of solution. What molar concentration of potassium chloride should appear on the label? (b) If the chemist had mistakenly used a 500 -mL. volumetric flask instead of the 150.0-mL flask in (a), what molar concentration of potassium chloride has the chemist actually prepared?

A chemist studying the propertics of photographic emulsions needed to prepare \(25.00 \mathrm{~mL}\). of \(0.155 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) (aq). What mass of silver nitrate must be placed into a \(25.00-\mathrm{mL}\) volumctric flask and dissolved and diluted to the mark with water?

(a) A 12.56-mL sample of \(1.345 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{~mL}\). What is the molar concentration of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the diluted solution? (b) A \(25.00-\mathrm{mL}\) sample of \(0.366 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is drawn from a reagent bottle with a pipet. The sample is transferred to a \(125.00-\mathrm{mL}\) volumetric flask and diluted to the mark with water. What is the molar concentration of the dilute hydrochloric acid solution?

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