Question

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" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) 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

Short answer

a. Mix 0.0278 L of concentrated H₂SO₄ (18 M) with enough water to make a total solution volume of 1.00 L. b. Mix 0.0417 L of concentrated HCl (12 M) with enough water to make a total solution volume of 1.00 L. c. Dissolve 118.85 g of NiCl₂⋅6H₂O in enough water to make a total solution volume of 1.00 L. d. Mix 0.0313 L of concentrated HNO₃ (16 M) with enough water to make a total solution volume of 1.00 L. e. Dissolve 53.00 g of Na₂CO₃ in enough water to make a total solution volume of 1.00 L.

Step-by-step solution

a. Preparing H₂SO₄ solution

We are given "concentrated" (18 M) sulfuric acid and we are supposed to prepare a 1.00 L (0.50 M) solution of it. To find the volume of concentrated sulfuric acid needed, we will use the M1 * V1 = M2 * V2 formula: M1 = 18 M (concentration of the concentrated sulfuric acid) V1 = volume of concentrated sulfuric acid needed M2 = 0.50 M (desired concentration of the solution to be prepared) V2 = 1.00 L (desired volume of the solution to be prepared) (18M) * V1 = (0.50M) * (1.00L) Now, solve for V1: V1 = (0.50M * 1.00L) / 18M = 0.0278 L So, in order to prepare the 1.00L of a 0.50M H₂SO₄ solution, we will need to mix 0.0278 L of concentrated sulfuric acid (18 M) with enough water to make total solution volume 1.00 L.

b. Preparing HCl solution

We are given "concentrated" (12 M) hydrochloric acid and we are supposed to prepare a 1.00 L (0.50 M) solution of it. Using the M1 * V1 = M2 * V2 formula: M1 = 12 M (concentration of the concentrated hydrochloric acid) V1 = volume of concentrated hydrochloric acid needed M2 = 0.50 M (desired concentration of the solution to be prepared) V2 = 1.00 L (desired volume of the solution to be prepared) (12M) * V1 = (0.50M) * (1.00L) Now, solve for V1: V1 = (0.50M * 1.00L) / 12M = 0.0417 L So, to prepare the 1.00L of a 0.50M HCl solution, we will need to mix 0.0417 L of concentrated hydrochloric acid (12 M) with enough water to make total solution volume 1.00 L.

c. Preparing NiCl₂ solution

We have to prepare a 1.00 L (0.50 M) solution of NiCl₂ from the salt NiCl₂⋅6H₂O. First, calculate the number of moles of NiCl₂ needed for a 0.50 M solution using the formula n2 = M2 * V2: n2 = number of moles of NiCl₂ needed M2 = 0.50 M (desired concentration of the solution) V2 = 1.00 L (desired volume of the solution) n2 = (0.50M) * (1.00L) = 0.50 mol Now, using the molar mass of NiCl₂⋅6H₂O, calculate the mass of the salt needed: The molar mass of NiCl₂⋅6H₂O = (58.69 + 2 * 35.45 + 6 * 18.02) g/mol = 237.69 g/mol mass = number of moles * molar mass = 0.50 mol * 237.69 g/mol = 118.85 g To prepare a 1.00L of a 0.50M NiCl₂ solution, we will need to dissolve 118.85 g of NiCl₂⋅6H₂O in enough water to make the total solution volume 1.00 L.

d. Preparing HNO₃ solution

We are given "concentrated" (16 M) nitric acid and we are supposed to prepare a 1.00 L (0.50 M) solution of it. Using the M1 * V1 = M2 * V2 formula: M1 = 16 M (concentration of the concentrated nitric acid) V1 = volume of concentrated nitric acid needed M2 = 0.50 M (desired concentration of the solution to be prepared) V2 = 1.00 L (desired volume of the solution to be prepared) (16M) * V1 = (0.50M) * (1.00L) Now, solve for V1: V1 = (0.50M * 1.00L) / 16M = 0.0313 L So, to prepare the 1.00L of a 0.50M HNO₃ solution, we will need to mix 0.0313 L of concentrated nitric acid (16 M) with enough water to make total solution volume 1.00 L.

e. Preparing Sodium Carbonate Solution

We have to prepare a 1.00 L (0.50 M) solution of sodium carbonate (Na₂CO₃) from the pure solid. First, calculate the number of moles of Na₂CO₃ needed for a 0.50 M solution using the formula n2 = M2 * V2: n2 = number of moles of Na₂CO₃ needed M2 = 0.50 M (desired concentration of the solution) V2 = 1.00 L (desired volume of the solution) n2 = (0.50M) * (1.00L) = 0.50 mol Now, using the molar mass of Na₂CO₃, calculate the mass of the salt needed: The molar mass of Na₂CO₃ = (2 * 22.99 + 12.01 + 3 * 16.00) g/mol = 105.99 g/mol mass = number of moles * molar mass = 0.50 mol * 105.99 g/mol = 53.00 g To prepare the 1.00L of a 0.50 M Na₂CO₃ solution, we will need to dissolve 53.00 g of sodium carbonate in enough water to make the total solution volume 1.00 L.

Key concepts

Understanding Molarity Concentration

Molarity concentration, denoted by the letter M, is a vital concept in chemistry that signifies the concentration of a solute within a solution. Essentially, it explains how much of a substance is present in a specific volume of solution. The formula for molarity is defined as the number of moles of solute divided by the volume of the solution in liters.

For instance, a molarity of 1 M means there is one mole of solute per liter of solution. To calculate it, use the equation: \[\begin{equation}M = \frac{n}{V}\end{equation}\]where M is the molarity, n is the number of moles of solute, and V is the volume of the solution in liters. In the context of the textbook exercise, preparing a 0.50 M HCl solution involves finding out the volume of concentrated HCl (known molarity of 12 M) needed to dilute it to the desired lower molarity.

The Process of Dilution of Solutions

Dilution is the process by which a solution's concentration is reduced by adding more solvent. The solute amount remains constant during this process, which is why the concept of dilution can be captured with the simple phrase 'dilution is the solution to pollution'. One of the most common formulations to understand dilution is the use of the formula: \[\begin{equation}M1 \cdot V1 = M2 \cdot V2\end{equation}\]where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume after dilution. Remember, as the volume (V) increases, molarity (M) decreases, provided the amount of solute does not change. In our exercise, to prepare a diluted solution, the volume (V1) of the concentrated solution required can be determined by rearranging the equation to solve for V1, while M2 and V2 represent the molarity and volume of the final diluted solution.

Stoichiometry of Reactions

Stoichiometry plays a crucial role in solution preparation and is rooted in the nature of chemical reactions. It deals with the quantitative relationships between the reactants and products in a chemical reaction. Stoichiometry is based on the conservation of mass where the mass of the reactants equals the mass of the products.

For preparing solutions from solid compounds, as in the case of NiCl₂⋅6H₂O and Na₂CO₃, stoichiometry is used to calculate the amount of substance (in moles) required to achieve a particular molarity. This calculation often begins with the desired molarity and volume of the final solution to find the number of moles needed:\[\begin{equation}n = M \cdot V\end{equation}\]Afterwards, this mole amount is converted to a mass using the molar mass of the compound, which entails multiplication with the formula weight expressed in grams per mole. This calculation ensures the precise amount of a substance is used to prepare a solution with the desired concentration.