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Describe how you would prepare 2.00 L of each of the following solutions. a. 0.250 M NaOH from solid NaOH b. 0.250 M NaOH from 1.00 M NaOH stock solution c. 0.100\(M \mathrm{K}_{2} \mathrm{CrO}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. 0.100\(M \mathrm{K}_{2} \mathrm{CrO}_{4}\) from 1.75 \(\mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

Short Answer

Expert verified
a. Dissolve 20.00 g of solid NaOH in water and dilute to a final volume of 2.00 L in a volumetric flask. b. Measure 0.500 L of the 1.00 M NaOH stock solution and dilute it to a final volume of 2.00 L in a volumetric flask. c. Dissolve 38.84 g of solid K2CrO4 in water and dilute to a final volume of 2.00 L in a volumetric flask. d. Measure 0.114 L of the 1.75 M K2CrO4 stock solution and dilute it to a final volume of 2.00 L in a volumetric flask.

Step by step solution

01

Calculate moles of NaOH

For a 2.00 L solution and a 0.250 M concentration, we will Calculate moles of NaOH using the given formula: \(moles = M \times V\) moles of NaOH = \((0.250 M) \times (2.00 L) = 0.500\) moles
02

Calculate the mass of NaOH

The molar mass of NaOH is 40.00 g/mol. To find the mass of NaOH needed: mass of NaOH = \((0.500\, moles) \times (40.00\, g/mol) = 20.00 g\)
03

Prepare the solution

Dissolve 20.00 g of solid NaOH in water and dilute to a final volume of 2.00 L in a volumetric flask. b. 0.250 M NaOH from 1.00 M NaOH stock solution
04

Calculate moles of NaOH

Same as in part a, we have 0.500 moles of NaOH.
05

Calculate the volume of stock solution

Using the stock solution concentration to find the volume needed: volume of stock solution = \(\frac{0.500\, moles}{1.00 M} = 0.500 L\)
06

Prepare the solution

Measure 0.500 L of the 1.00 M NaOH stock solution and dilute it to a final volume of 2.00 L in a volumetric flask. c. 0.100 M K2CrO4 from solid K2CrO4
07

Calculate moles of K2CrO4

For a 2.00 L solution and a 0.100 M concentration: moles of K2CrO4 = \((0.100 M) \times (2.00 L) = 0.200\) moles
08

Calculate the mass of K2CrO4

The molar mass of K2CrO4 is 194.19 g/mol. To find the mass of K2CrO4 needed: mass of K2CrO4 = \((0.200\, moles) \times (194.19\, g/mol) = 38.84 g\)
09

Prepare the solution

Dissolve 38.84 g of solid K2CrO4 in water and dilute to a final volume of 2.00 L in a volumetric flask. d. 0.100 M K2CrO4 from 1.75 M K2CrO4 stock solution
10

Calculate moles of K2CrO4

Same as in part c, we have 0.200 moles of K2CrO4.
11

Calculate the volume of stock solution

Using the stock solution concentration to find the volume needed: volume of stock solution = \(\frac{0.200\, moles}{1.75 M} = 0.114 L\)
12

Prepare the solution

Measure 0.114 L of the 1.75 M K2CrO4 stock solution and dilute it to a final volume of 2.00 L in a volumetric flask.

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Key Concepts

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

Molarity Calculations
Molarity (\(M\)) is a way to express concentration, showing the number of moles of solute per liter of solution. To calculate molarity, use the formula:\[M = \frac{moles\,of\,solute}{liters\,of\,solution}\]A key step is understanding how to find moles of a particular solute. For example, to prepare a 0.250 M NaOH solution, we first need to determine how many moles the solution needs. Given 2.00 liters (L) at 0.250 M, use:
  • \(moles = M \times V = 0.250 \times 2.00 = 0.500\) moles.
Once you've determined the moles, you can either calculate the mass of a solid solute required or the volume of a stock solution needed.
Dilution from Stock Solutions
Dilution involves reducing the concentration of a solution, usually by adding more solvent. To dilute from a stock solution, you can use the formula:\[V_{1} \times M_{1} = V_{2} \times M_{2}\]where \( V_{1} \) and \( M_{1} \) are the volume and molarity of the initial solution, and \( V_{2} \) and \( M_{2} \) are those of the final solution.For example, to prepare 2.00 L of 0.250 M NaOH from a 1.00 M stock solution, find the initial volume:
  • \( V_{1} = \frac{0.500\,moles}{1.00\,M} = 0.500\,L\).
Measure 0.500 L of stock, then add more water to reach the final 2.00 L.
Molar Mass Calculations
The molar mass of a substance is the weight in grams of one mole of that substance and is crucial for converting between mass and moles. You determine it by adding the atomic masses of all atoms in a molecule, found on the periodic table.To convert moles to grams, use the calculation:\[mass = moles \times molar\,mass\]For example, NaOH has a molar mass of 40.00 g/mol. For 0.500 moles, calculate mass needed:
  • \( mass = 0.500 \times 40.00 = 20.00 \) grams.
This guides how much solid NaOH you'll dissolve to make the solution.

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Most popular questions from this chapter

A mixture contains only sodium chloride and potassium chloride. A 0.1586-g sample of the mixture was dissolved in water. It took 22.90 mL of 0.1000 M AgNO3 to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?

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Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

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