A student investigating the propertics of solutions containing carbonate ions prepared a solution containing \(7.112 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in a \(250.0-\mathrm{mL}\) volumetric flask. Some of the solution was transferred to a burer. What volume of solution should be dispensed from the buret to provide (a) \(5.112 \times 10^{-3} \mathrm{~mol} \mathrm{Na}_{2} \mathrm{CO}_{3}\); (b) \(3.451 \times 10^{-3} \mathrm{~mol} \mathrm{CO}_{3}^{2-}\) ?

Short Answer

Expert verified
The volumes to dispense are (a) the volume for 5.112 x 10^-3 mol Na2CO3 and (b) the volume for 3.451 x 10^-3 mol CO32-, calculated using the molarity of the prepared Na2CO3 solution.

Step by step solution

01

Calculate the molar mass of Na2CO3

To find the molar mass of sodium carbonate (Na2CO3), add the molar masses of all atoms in the formula: (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O), which is approximately (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g/mol.
02

Find the molarity of the Na2CO3 solution

Molarity (M) is the number of moles of solute per liter of solution. Since 7.112 g Na2CO3 is dissolved in 250.0 mL (0.2500 L), first find the moles of Na2CO3 using its molar mass. Moles of Na2CO3 = 7.112 g / 105.99 g/mol. Then calculate the molarity: Molarity = Moles of Na2CO3 / Volume of solution in liters.
03

Calculate the volume needed to dispense 5.112 x 10^-3 mol Na2CO3

Use the known molarity (M) of the solution to find the volume (V) with the equation n = MV, where n is the moles of solute. Rearrange to find V: V = n / M. Substitute the number of moles (5.112 x 10^-3 mol) and the molarity from Step 2 to solve for V.
04

Calculate the volume needed to dispense 3.451 x 10^-3 mol CO32-

One mole of Na2CO3 contains one mole of CO32-. Therefore, the molarity of CO32- ions is the same as the molarity of Na2CO3 in the solution. Use the molarity and the moles of CO32- (3.451 x 10^-3 mol) with the equation V = n / M to solve for V.

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Key Concepts

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

Stoichiometry
Stoichiometry is the mathematical relationship between the quantities of reactants and products in chemical reactions, including the masses of substances and the volume of solutions involved.

To solve stoichiometric problems, the first step is often to balance the chemical equation to ensure the law of conservation of mass is respected. Once the equation is balanced, the mole ratio of reactants to products can be used to convert between masses and moles, or to deduce the volume of a solution containing a certain amount of substance.

To tackle exercises like calculating the volume of solution needed to provide a certain number of moles of a substance, a solid understanding of stoichiometry is essential. This generally involves determining molar ratios, converting grams to moles (or vice versa), and applying the concept of molarity as a measure of concentration.
Molar Mass
The molar mass of a substance is the weight of one mole of that substance, typically expressed in grams per mole (g/mol). It's a pivotal value in chemistry for converting between the mass of a substance and the amount of substance in moles.

The molar mass is the sum of the atomic masses of all the atoms in a molecule's formula. In the context of the provided exercise, calculating the molar mass of sodium carbonate (a_2CO_3) is crucial for further molarity calculations. This process involves using the periodic table to find the atomic masses of sodium (Na), carbon (C), and oxygen (O), and then summing these values according to the molecule's composition.
Concentration of Solutions
When dealing with solutions, the concentration tells us how much solute is contained in a specific volume of solvent. Molarity (M), defined as the number of moles of solute per liter of solution, is a commonly used unit of concentration in chemistry.

Understanding the concentration of a solution becomes vital when preparing solutions with precise molarities or when performing titration experiments, as seen in the exercise. Knowledge of molarity allows you to calculate the exact volume of solution necessary to achieve a desired number of moles of a solute, which is often required in quantitative analysis and laboratory preparations. This exercise demonstrates how the molarity concept is applied to calculate the volume of an a_2CO_3 solution needed to dispense a specified amount of solute.
Analytical Chemistry
Analytical chemistry is the branch of chemistry that involves the analysis of material samples to understand their chemical composition and structure. Techniques in analytical chemistry, such as titration, allow scientists and students to quantify the concentration of a solute in a solution.

In the context of the exercise provided, the ability to calculate the molarity of a solution and then use it to find out how much of the solution must be dispensed from a buret directly ties into analytical chemistry practices. Such calculations are fundamental in preparing solutions with precise concentrations, which are indispensable for accurate and reliable chemical analysis and experiments. The practice of these techniques strengthens understanding of key analytical concepts crucial to the field.

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Most popular questions from this chapter

Calculate the molality of the solute in each of the following solutions: (a) \(10.0 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolvcd in \(250 \mathrm{~g}\) water; (b) \(0.48 \mathrm{~mol}\) of \(\mathrm{KOH}\) dissolved in \(50.0 \mathrm{~g}\) water; (c) \(1.94 \mathrm{~g}\) of urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), dissolved in \(200 \mathrm{~g}\) water.

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.)

Describe the preparation of cach solution, starting with the anhydrous solute and warer and using the corrcsponding volumetric flask: (a) \(25.0 \mathrm{~mL}\) of \(6.0 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) (b) \(1.0 \mathrm{~L}\), of \(0.10 \mathrm{M} \mathrm{BaCl}_{2}\) (aq); (c) \(500 \mathrm{~mL}\) of \(0.0010 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\).

(a) Determine the mass of anhydrous copper(II) sulfate that must be used to prepare \(250 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution. (b) Determine the mass of \(\mathrm{CuSO}_{4}-5 \mathrm{H}_{2} \mathrm{O}\) that must be used to prepare \(250 \mathrm{ml}\). of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution.

Calculate the mole fraction of each component in the following solutions: (a) \(2.5 .0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\); (b) \(25.0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\); (c) a glucowe solution that is \(0.10 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\).

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