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Chapter 4: Problem 9

You need to make \(150.0 \mathrm{~mL}\) of a \(0.10-M \mathrm{NaCl}\) solution. You have solid \(\mathrm{NaCl}\), and your lab partner has a \(2.5-\mathrm{M} \mathrm{NaCl}\) solution. Explain how you each make the \(0.10-M \mathrm{NaCl}\) solution.

Short Answer

Expert verified
To make a \(0.10-M \mathrm{NaCl}\) solution using solid \(\mathrm{NaCl}\), dissolve \(0.877 \, \mathrm{g}\) of solid \(\mathrm{NaCl}\) in \(150 \mathrm{~mL}\) of distilled water. To make the same solution using the given \(2.5-M \mathrm{NaCl}\) solution, dilute \(6.0 \mathrm{~mL}\) of the \(2.5-M \mathrm{NaCl}\) solution with distilled water in a \(150 \mathrm{~mL}\) volumetric flask.

Step by step solution

01

Preparing the \(0.10-M \mathrm{NaCl}\) solution using solid \(\mathrm{NaCl}\).

1. Calculate the moles of \(\mathrm{NaCl}\) needed for the desired concentration and volume: For a \(0.10-M \mathrm{NaCl}\) solution, the formula is: Moles of \(\mathrm{NaCl}\) = volume of solution (L) 脳 molarity (\(M\)) First, convert \(150 \mathrm{~mL}\) to liters: \(150 \mathrm{~mL} = 0.15 \mathrm{~L}\). Moles of \(\mathrm{NaCl}\) = \(0.15 \mathrm{~L} \times 0.10-M = 0.015 \, \mathrm{mol}\). 2. Calculate the mass of solid \(\mathrm{NaCl}\) needed: Using the molar mass of \(\mathrm{NaCl}\): \(58.44 \, \mathrm{g/mol}\), Mass of \(\mathrm{NaCl}\) = moles of \(\mathrm{NaCl} \times\) molar mass Mass of \(\mathrm{NaCl} = 0.015 \, \mathrm{mol} \times 58.44 \,\mathrm{g/mol} = 0.877 \, \mathrm{g}\). 3. Preparing the solution: Weigh out \(0.877 \, \mathrm{g}\) of solid \(\mathrm{NaCl}\), and dissolve it in approximately \(100 \mathrm{~mL}\) of distilled water in a volumetric flask. Fill the volumetric flask to the \(150 \mathrm{~mL}\) mark with distilled water, and mix thoroughly to obtain the desired \(0.10-M \mathrm{NaCl}\) solution.
02

Preparing the \(0.10-M \mathrm{NaCl}\) solution using the given \(2.5-M \mathrm{NaCl}\) solution.

1. Calculate the volume of \(2.5-M \mathrm{NaCl}\) solution needed: Using dilution formula: \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume of the given solution, and \(C_2\) and \(V_2\) are the final concentration and volume of the desired solution. Rearrange the formula for \(V_1\): \(V_1 = \frac{C_2V_2}{C_1}\). 2. Calculate the required volume: Plug in the desired concentration \(C_2 = 0.10 \mathrm{~M}\), the desired volume \(V_2 = 0.15 \mathrm{~L}\), and the given concentration \(C_1 = 2.5 \mathrm{~M}\): \(V_1 = \frac{0.10-M \times 0.15 \mathrm{~L}}{2.5 \mathrm{~M}} = 0.006 \mathrm{~L} = 6.0 \mathrm{~mL}\). 3. Preparing the solution: Using a volumetric pipette, measure out \(6.0 \mathrm{~mL}\) of the \(2.5-M \mathrm{NaCl}\) solution, and transfer it to a \(150 \mathrm{~mL}\) volumetric flask. Fill the volumetric flask to the \(150 \mathrm{~mL}\) mark with distilled water, and mix thoroughly to obtain the desired \(0.10-M \mathrm{NaCl}\) solution.

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Key Concepts

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

Molarity calculations
Molarity is an important concept in chemistry that represents the concentration of a solution. It's simply the amount of solute (the substance being dissolved) divided by the volume of the solution in liters. For example, when preparing a 0.10-M NaCl solution, you need to calculate how many moles of NaCl are required for your desired volume.

To find the moles of NaCl, use the formula: \[\text{Moles of NaCl} = \text{volume of solution (L)} \times \text{molarity (M)}\]In the case of our solution, convert the volume from milliliters to liters first (\(150\text{ mL} = 0.15\text{ L}\)), and then plug it into the formula to get:\[0.15\text{ L} \times 0.10\text{ M} = 0.015\text{ mol}\]Knowing how to perform molarity calculations allows you to tailor solutions with precision, which is critical in both laboratory settings and industrial applications. It's like having a recipe to make sure your solutions are just right!
Dilution process
The dilution process is essential when you have a concentrated solution and need to create a less concentrated one. You use the dilution formula:\[C_1V_1 = C_2V_2\]Where:
  • \(C_1\) is the initial concentration
  • \(V_1\) is the initial volume
  • \(C_2\) is the final concentration
  • \(V_2\) is the final volume
This formula helps us determine how much of the concentrated solution is needed to achieve our desired concentration. In our scenario, to prepare a 0.10-M solution from a 2.5-M solution (provided by your lab partner), you rearrange the formula to solve for \(V_1\):\[V_1 = \frac{C_2V_2}{C_1}\]Substituting the values:\[V_1 = \frac{0.10\text{ M} \times 0.15\text{ L}}{2.5\text{ M}} = 0.006\text{ L} = 6.0\text{ mL}\]This calculation shows you need 6.0 mL of the 2.5-M solution diluted with water to make 150.0 mL of a 0.10-M solution.
Solution preparation steps
Preparing a solution requires a step-by-step approach to ensure accuracy and safety. When you're starting from solid NaCl, you'll perform the following steps:

  • Weigh the exact mass of NaCl needed. In our case, that鈥檚 0.877 grams because the molar mass of NaCl is 58.44 g/mol.
  • Dissolve this measured NaCl in a smaller volume of water, about 100 mL initially, to make sure it completely dissolves before reaching the final volume.
  • Transfer the solution to a volumetric flask and add water until you reach the desired total volume of 150 mL.
  • Mix thoroughly to ensure the solution is uniform.
When using an existing solution, like our 2.5-M NaCl solution, follow these steps instead:

  • Use a volumetric pipette to accurately measure 6.0 mL of the concentrated solution.
  • Transfer this to a 150 mL volumetric flask.
  • Add distilled water up to the 150 mL mark.
  • Mix thoroughly to achieve a uniform concentration.
These preparation steps are foundational in chemistry labs and ensure that your solutions are created with precision and appropriate concentration, making subsequent experiments reliable and accurate.

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Most popular questions from this chapter

In a \(1-\mathrm{L}\) beaker, \(203 \mathrm{~mL}\) of \(0.307 \mathrm{M}\) ammonium chromate was mixed with \(137 \mathrm{~mL}\) of \(0.269 M\) chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical equation for the reaction occurring here. If the percent yield of the reaction was \(88.0 \%\), what mass of chromium(III) chromate was isolated?

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl} ?\)

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0\) ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals \(1.0 \mathrm{~g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" ( \(18 \mathrm{M}\) ) sulfuric acid b. HCl from "concentrated" (12 \(M\) ) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" ( \(16 M\) ) reagent e. Sodium carbonate from the pure solid

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584\) g pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution \(A, 50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\). For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C .\)
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