Question

How would you prepare 1.00 L of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) 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_ from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

Short answer

To prepare 1.00 L of 0.50-M solutions, use the following volumes and masses: a. 27.8 mL of 18 M concentrated H₂SO₄ b. 41.7 mL of 12 M concentrated HCl c. 115.83 g of NiCl₂.6H₂O salt d. 31.2 mL of 16 M concentrated HNO₃ e. 52.99 g of sodium carbonate (Na₂CO₃) Measure the respective amounts, dissolve in appropriate volumes of distilled water, and mix well to create the desired solutions.

Step-by-step solution

a. Preparing 0.50-M H2SO4 solution from 18 M concentrated sulfuric acid

To prepare a 0.50-M solution, we first need to determine the volume of the concentrated acid to be used. We can use the formula M1V1 = M2V2, where M1 is the initial concentration of the acid, V1 is the volume of the concentrated acid required, M2 is the final concentration of the acid, and V2 is the final volume of the solution. Given, M1 = 18 M, M2 = 0.50 M, and V2 = 1.00 L, to find V1, we can rearrange the equation as follows: V1 = (M2 * V2) / M1 V1 = (0.50 mol/L * 1.00 L) / 18 mol/L = 0.0278 L Convert 0.0278 L to mL: V1 = 0.0278 * 1000 = 27.8 mL Now, to prepare 1.00 L of 0.50 M H2SO4 solution, follow these steps: 1. Measure 27.8 mL of 18 M concentrated H2SO4 using a graduated cylinder. 2. Transfer the measured H2SO4 into a 1 L volumetric flask. 3. Add distilled water to the flask, up to the 1 L mark, and mix well.

b. Preparing 0.50-M HCl solution from 12 M concentrated reagent

Given, M1 = 12 M, M2 = 0.50 M, and V2 = 1.00 L. Using the formula M1V1 = M2V2: V1 = (M2 * V2) / M1 V1 = (0.50 mol/L * 1.00 L) / 12 mol/L = 0.0417 L Convert 0.0417 L to mL: V1 = 0.0417 * 1000 = 41.7 mL To prepare 1.00 L of 0.50 M HCl solution: 1. Measure 41.7 mL of 12 M concentrated HCl using a graduated cylinder. 2. Transfer the measured HCl into a 1 L volumetric flask. 3. Add distilled water to the flask, up to the 1 L mark, and mix well.

c. Preparing 0.50-M NiCl2 solution from NiCl2.6H2O salt

First, find the molar mass of NiCl2.6H2O: Ni = 58.69 g/mol, Cl = 35.45 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol Molar mass of NiCl₂.6H₂O = Ni + 2Cl + (12H + 6O) = 58.69 + 2(35.45) + 6(1.01 + 16.00) = 58.69 + 70.90 + 102.06 = 231.65 g/mol Next, find the mass of NiCl2.6H2O required to make 1 L of 0.50-M solution: 1 L * 0.50 mol/L * 231.65 g/mol = 115.83 g To prepare 1.00 L of 0.50 M NiCl2 solution: 1. Measure 115.83 g of NiCl2.6H2O salt using a balance. 2. Dissolve the measured salt in a beaker with some distilled water. 3. Transfer the dissolved solution into a 1 L volumetric flask. 4. Add distilled water to the flask, up to the 1 L mark, and mix well.

d. Preparing 0.50-M HNO3 solution from 16 M concentrated reagent

Given, M1 = 16 M, M2 = 0.50 M, and V2 = 1.00 L. Using the formula M1V1 = M2V2: V1 = (M2 * V2) / M1 V1 = (0.50 mol/L * 1.00 L) / 16 mol/L = 0.0312 L Convert 0.0312 L to mL: V1 = 0.0312 * 1000 = 31.2 mL To prepare 1.00 L of 0.50 M HNO3 solution: 1. Measure 31.2 mL of 16 M concentrated HNO3 using a graduated cylinder. 2. Transfer the measured HNO3 into a 1 L volumetric flask. 3. Add distilled water to the flask, up to the 1 L mark, and mix well.

e. Preparing 0.50-M Sodium Carbonate solution from the pure solid

First, find the molar mass of sodium carbonate (Na2CO3): Na = 22.99 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol Molar mass of Na₂CO₃ = 2Na + C + 3O = 2(22.99) + 12.01 + 3(16.00) = 45.98 + 12.01 + 48.00 = 105.99 g/mol Next, find the mass of Na2CO3 required to make 1 L of 0.50-M solution: 1 L * 0.50 mol/L * 105.99 g/mol = 52.99 g To prepare 1.00 L of 0.50 M sodium carbonate solution: 1. Measure 52.99 g of sodium carbonate (Na2CO3) using a balance. 2. Dissolve the measured salt in a beaker with some distilled water. 3. Transfer the dissolved solution into a 1 L volumetric flask. 4. Add distilled water to the flask, up to the 1 L mark, and mix well.

Key concepts

Molarity Calculation

Molarity is a measurement of concentration, representing the number of moles of a solute per liter of solution. Knowing how to calculate molarity is essential in chemistry, especially when preparing solutions.

Molarity, denoted by the symbol 'M', is calculated using the formula: \[ M = \frac{{n}}{{V}} \]where 'n' is the number of moles of the solute and 'V' is the volume of the solution in liters. To find the molarity of a solution, we can rearrange this formula to solve for the unknown quantity, as long as we have the other two.

In the practice of preparing solutions, a common task is to dilute a concentrated solution to attain a desired molarity. The dilution equation, \[ M_1V_1 = M_2V_2 \], is vital. This equation means that the product of the original concentration (\( M_1 \)) and the original volume (\( V_1 \)) of the solution equals the product of the final concentration (\( M_2 \)) and the final volume (\( V_2 \)). It's widely used to calculate how much of the concentrated stock solution is needed to achieve a specific concentration and volume of a new solution.

Solution Dilution

When working with chemical solutions, sometimes we need to lower the concentration by adding more solvent, a process known as dilution. The concept of dilution is fundamental in labs for achieving desired solution concentrations without having to prepare a fresh batch every time.

To dilute a solution, one simply takes a measure of the stock solution and adds solvent (commonly water) until reaching the desired concentration. The molarity equation \( M_1V_1 = M_2V_2 \) provides a straightforward method to determine how much of the stock solution is needed for a given dilution. In essence, you are keeping the amount of substance constant (moles), while increasing the volume.

It is important to acknowledge that while diluting, the 'solute' remains the same, only the volume of 'solvent' changes. Thus, understanding dilution enables one to prepare solutions of accurate molarity efficiently, which is crucial for experimental consistencies and cost-effectiveness in a laboratory setting.

Molar Mass Determination

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a physical property that is used for many calculations in chemistry, such as when converting between mass and moles of a substance.

To determine the molar mass, one needs to sum the atomic masses of all the atoms in the compound. Each element's atomic mass is found on the periodic table, and for a molecule, the total molar mass is the sum of the molar masses of each individual element multiplied by the number of atoms of that element in the molecule. For hydrates, water molecules associated with the compound must also be included in the calculation.

For example, in the exercise, the molar mass of \( NiCl_2 \bullet 6H_2O \) was calculated by adding the molar masses of nickel (Ni), chlorine (Cl), hydrogen (H), and oxygen (O) corresponding to their molecular formula. The calculation included the atoms from the six water molecules (\( 6H_2O \)) that are part of the crystal structure. Determining the correct molar mass is vital as it directly impacts the preparation of solutions with accurate molarity.