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Magazine Chapter 4: Problem 42
How would you prepare 1.00 L of a 0.50-M solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) 鈥渃oncentrated鈥� (18 M) sulfuric acid b. HCl from 鈥渃oncentrated鈥� (12 M) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{HNO}_{3}\) from 鈥渃oncentrated鈥� (16 M) reagent e. Sodium carbonate from the pure solid
Short Answer
Expert verified
To prepare 1.00 L of a 0.50-M solution of the following compounds:
a. Add 27.8 mL of 18 M H鈧係O鈧� to 1 L of distilled water.
b. Add 41.7 mL of 12 M HCl to 1 L of distilled water.
c. Dissolve 118.86 g of NiCl鈧偮�6H鈧侽 in 1 L of distilled water.
d. Add 31.3 mL of 16 M HNO鈧� to 1 L of distilled water.
e. Dissolve 52.995 g of Na鈧侰O鈧� in 1 L of distilled water.
Step by step solution
01
Identify the initial concentration and desired volume
For concentrated H2SO4, the initial concentration M1=18 M. In this case, we want to prepare a 0.50-M solution (M2) with a final volume V2=1.00 L.
02
Determine the volume of concentrated acid needed
Apply the dilution formula:
\[M_1V_1 = M_2V_2\]
\[V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50 \,\text{M})(1.00\, \text{L})}{18 \,\text{M}} = 0.0278\, \text{L}\]
03
Dilute the concentrated acid
Measure 0.0278 L (27.8 mL) of concentrated H2SO4 and slowly add it to a volumetric flask already containing approximately 0.5 L of distilled water. Mix well and then add more distilled water until the total volume is 1.00 L.
#b. Preparing 0.50-M HCl from 12-M reagent#
Follow the same steps as for H2SO4.
04
Identify the initial concentration and desired volume
For concentrated HCl, the initial concentration M1=12 M.
05
Determine the volume of concentrated reagent needed
Apply the dilution formula:
\[V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50 \,\text{M})(1.00\, \text{L})}{12 \,\text{M}} = 0.0417\, \text{L}\]
06
Dilute the concentrated reagent
Measure 0.0417 L (41.7 mL) of concentrated HCl and slowly add it to a volumetric flask already containing approximately 0.5 L of distilled water. Mix well and then add more distilled water until the total volume is 1.00 L.
#c. Preparing 0.50-M NiCl2 from the salt NiCl2路6H2O#
07
Calculate the required amount of salt (in moles)
Since M2 = 0.50 M and V2 = 1.00 L, we need 0.50 moles of NiCl2.
08
Convert moles to grams
Calculate the molar mass of NiCl2路6H2O:
\[1\, \text{Ni}: 58.69 \, g/mol\]
\[2\, \text{Cl}: 2(35.45) = 70.90 \, g/mol\]
\[12\, \text{H}: 12(1.01) = 12.12 \, g/mol\]
\[6\, \text{O}: 6(16.00) = 96.00 \, g/mol\]
\[Total = 237.71\, g/mol\]
Now, multiply the number of moles by the molar mass:
\[(0.50\, \text{mol})(237.71\, g/mol) = 118.86\, g\]
09
Dissolve the salt to prepare the solution
Weigh 118.86 g of the NiCl2路6H2O and dissolve it in a volumetric flask already containing approximately 0.5 L of distilled water. Mix well and then add more distilled water until the total volume is 1.00 L.
#d. Preparing 0.50-M HNO3 from 16-M reagent#
Follow the same steps as for H2SO4.
10
Identify the initial concentration and desired volume
For concentrated HNO3, the initial concentration M1=16 M.
11
Determine the volume of concentrated reagent needed
Apply the dilution formula:
\[V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50 \,\text{M})(1.00\, \text{L})}{16 \,\text{M}} = 0.0313\, \text{L}\]
12
Dilute the concentrated reagent
Measure 0.0313 L (31.3 mL) of concentrated HNO3 and slowly add it to a volumetric flask already containing approximately 0.5 L of distilled water. Mix well and then add more distilled water until the total volume is 1.00 L.
#e. Preparing 0.50-M Sodium carbonate from the pure solid#
Follow the same steps as for NiCl2路6H2O.
13
Calculate the required amount of solid (in moles)
Since M2 = 0.50 M and V2 = 1.00 L, we need 0.50 moles of sodium carbonate.
14
Convert moles to grams
Calculate the molar mass of sodium carbonate, Na2CO3:
\[2\, \text{Na}: 2(22.99) = 45.98 \, g/mol\]
\[1\, \text{C}: 12.01 \, g/mol\]
\[3\, \text{O}: 3(16.00) = 48.00 \, g/mol\]
\[Total = 105.99\, g/mol\]
Now, multiply the number of moles by the molar mass:
\[(0.50\, \text{mol})(105.99\, g/mol) = 52.995\, g\]
15
Dissolve the solid to prepare the solution
Weigh 52.995 g of sodium carbonate and dissolve it in a volumetric flask already containing approximately 0.5 L of distilled water. Mix well and then add more distilled water until the total volume is 1.00 L.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
Molarity is an essential concept in chemistry, referring to the concentration of a solution. It shows the number of moles of solute per liter of solution. This is depicted with the formula, \( M = \frac{n}{V} \), where \( M \) represents molarity, \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters. This measurement helps chemists and scientists understand how much solute is present in a given volume of solvent, making it easier to predict chemical reactions and behavior.
When preparing a solution, knowing the molarity allows for precise mixing and consistency. For example, to make a 0.50-M solution, you need to ensure that for every liter of solution, there is exactly 0.50 moles of the solute present. This precision is critical in scientific experiments, where even small deviations can have significant effects.
When preparing a solution, knowing the molarity allows for precise mixing and consistency. For example, to make a 0.50-M solution, you need to ensure that for every liter of solution, there is exactly 0.50 moles of the solute present. This precision is critical in scientific experiments, where even small deviations can have significant effects.
Dilution
Dilution is a process used to decrease the concentration of a solution by adding more solvent. The relationship between the initial and final states of the solution can be described by the dilution formula: \( M_1V_1 = M_2V_2 \). This formula helps to calculate the amount of concentrated solution needed to achieve a desired molarity after dilution.
For example, if you have a 12 M (molar) concentrated solution of hydrochloric acid (HCl) and need to prepare a 0.50 M solution, you have to calculate how much concentrated solution is needed to mix with water to reach the desired concentration. Calculating using the formula ensures the accuracy of the solution's new concentration. This is vital in various applications, such as chemical manufacturing and laboratory experiments.
For example, if you have a 12 M (molar) concentrated solution of hydrochloric acid (HCl) and need to prepare a 0.50 M solution, you have to calculate how much concentrated solution is needed to mix with water to reach the desired concentration. Calculating using the formula ensures the accuracy of the solution's new concentration. This is vital in various applications, such as chemical manufacturing and laboratory experiments.
Volumetric Flask
A volumetric flask is a type of laboratory glassware used for preparing solutions of precise volumes. It is typically pear-shaped with a flat bottom, and there is a long neck marked with a line. This mark indicates the exact volume that the flask can hold.
Volumetric flasks are crucial in the preparation of solutions due to their accuracy. They are designed to contain a precise amount of liquid at a particular temperature, which is significant in chemical experiments where solution concentration needs to be exact. When preparing a solution, you would fill the volumetric flask with distilled water until it's half full, add the measured amount of solute, and then fill to the mark with more distilled water. This ensures the solvent and solute are evenly distributed throughout the flask.
Volumetric flasks are crucial in the preparation of solutions due to their accuracy. They are designed to contain a precise amount of liquid at a particular temperature, which is significant in chemical experiments where solution concentration needs to be exact. When preparing a solution, you would fill the volumetric flask with distilled water until it's half full, add the measured amount of solute, and then fill to the mark with more distilled water. This ensures the solvent and solute are evenly distributed throughout the flask.
Molar Mass
Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It is used to convert moles of a substance into grams, which is often more practical for laboratory use.
Calculating molar mass involves summing the masses of the atomic constituents of a compound based on their individual atomic masses found in the periodic table. For instance, for nickel chloride hexahydrate, \( \text{NiCl}_2 \cdot 6 \text{H}_2\text{O} \), the molar mass is calculated by adding the molar masses of all atoms in the formula.
Calculating molar mass involves summing the masses of the atomic constituents of a compound based on their individual atomic masses found in the periodic table. For instance, for nickel chloride hexahydrate, \( \text{NiCl}_2 \cdot 6 \text{H}_2\text{O} \), the molar mass is calculated by adding the molar masses of all atoms in the formula.
- Nickel (Ni): 58.69 g/mol
- Chlorine (Cl): 35.45 g/mol, multiplied by 2
- Hydrogen (H): 1.01 g/mol, multiplied by 12
- Oxygen (O): 16.00 g/mol, multiplied by 6
