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Chapter 4: Problem 42

How would you prepare 1.00 L of a 0.50-M solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) “concentrated” (18 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. \(\mathrm{HNO}_{3}\) from “concentrated” (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 H₂SO₄ 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 NiCl₂·6H₂O in 1 L of distilled water. d. Add 31.3 mL of 16 M HNO₃ to 1 L of distilled water. e. Dissolve 52.995 g of Na₂CO₃ 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 NiCl2·6H2O#
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 NiCl2·6H2O: \[1\, \text{Ni}: 58.69 \, g/mol\] \[2\, \text{Cl}: 2(35.45) = 70.90 \, g/mol\] \[12\, \text{H}: 12(1.01) = 12.12 \, g/mol\] \[6\, \text{O}: 6(16.00) = 96.00 \, g/mol\] \[Total = 237.71\, g/mol\] 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 NiCl2·6H2O 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 NiCl2·6H2O.
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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