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" ( \(18 \mathrm{M}\) ) 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

Short answer

To prepare 1.00 L of 0.50 M solutions: a. Measure 27.8 mL of 18 M H₂SO₄ and dilute with water to 1.00 L. b. Measure 41.7 mL of 12 M HCl and dilute with water to 1.00 L. c. Dissolve 118.85 g of NiCl₂·6H₂O in water to reach 1.00 L. d. Measure 31.3 mL of 16 M HNO₃ and dilute with water to 1.00 L. e. Dissolve 52.995 g of sodium carbonate in water to reach 1.00 L.

Step-by-step solution

Identify the initial and final concentrations and volumes

We are given the initial concentration (\(M_1\)) of the concentrated sulfuric acid as 18 M, and the desired final concentration (\(M_2\)) as 0.50 M. The final volume (\(V_2\)) should be 1.00 L.

Apply the dilution formula

Using the dilution formula, \(M_1V_1 = M_2V_2\), we can find the initial volume (\(V_1\)) required to prepare the final solution. \(V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50\,\text{M})(1.00\,\text{L})}{18\,\text{M}} = 0.0278\,\text{L}\)

Calculate and perform the dilution

We need to measure 0.0278 L (27.8 mL) of concentrated sulfuric acid and dilute it with enough water to reach a final volume of 1.00 L. #b. Preparing 1.00 L of 0.50 M HCl from 12 M Concentrated Reagent#

Identify the initial and final concentrations and volumes

The initial concentration (\(M_1\)) of the concentrated HCl solution is 12 M, and the desired final concentration (\(M_2\)) is 0.50 M. The final volume (\(V_2\)) should be 1.00 L.

Apply the dilution formula

Using the dilution formula, \(M_1V_1 = M_2V_2\), we can find the initial volume (\(V_1\)) required to prepare the final solution. \(V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50\,\text{M})(1.00\,\text{L})}{12\,\text{M}} = 0.0417\,\text{L}\)

Calculate and perform the dilution

We need to measure 0.0417 L (41.7 mL) of concentrated HCl and dilute it with enough water to reach a final volume of 1.00 L. #c. Preparing 1.00 L of 0.50 M NiCl₂ from NiCl₂·6H₂O salt#

Calculate the moles of solute needed

To prepare the final solution, we need to find the moles of NiCl₂ needed using the formula \(n =MV\). Here, n is the number of moles, M is the molarity (0.50 M), and V is the volume (1.00 L). \(n = (0.50\,\text{M})(1.00\,\text{L}) = 0.50\,\text{mol}\)

Convert moles to mass using molar mass

We will now convert moles of NiCl₂·6H₂O to mass using the molar mass of the compound, which is approximately 237.69 g/mol. \(m = n \cdot \text{MM} = (0.50\,\text{mol})(237.69\,\text{g/mol}) = 118.85\,\text{g}\)

Dissolve the salt in water

Weigh 118.85 g of NiCl₂·6H₂O and dissolve it in enough water to reach a final volume of 1.00 L. #d. Preparing 1.00 L of 0.50 M HNO₃ from 16 M Concentrated Reagent#

Identify the initial and final concentrations and volumes

The initial concentration (\(M_1\)) of the concentrated HNO₃ solution is 16 M, and the desired final concentration (\(M_2\)) is 0.50 M. The final volume (\(V_2\)) should be 1.00 L.

Apply the dilution formula

Using the dilution formula, \(M_1V_1 = M_2V_2\), we can find the initial volume (\(V_1\)) required to prepare the final solution. \(V_1 = \frac{M_2V_2}{M_1} = \frac{(0.50\,\text{M})(1.00\,\text{L})}{16\,\text{M}} = 0.0313\,\text{L}\)

Calculate and perform the dilution

We need to measure 0.0313 L (31.3 mL) of concentrated HNO₃ and dilute it with enough water to reach a final volume of 1.00 L. #e. Preparing 1.00 L of 0.50 M Sodium Carbonate from the Pure Solid#

Calculate the moles of solute needed

To prepare the final solution, we need to find the moles of sodium carbonate needed using the formula \(n =MV\). Here, n is the number of moles, M is the molarity (0.50 M), and V is the volume (1.00 L). \(n = (0.50\,\text{M})(1.00\,\text{L}) = 0.50\,\text{mol}\)

Convert moles to mass using molar mass

We will now convert moles of sodium carbonate to mass using the molar mass of the compound, which is approximately 105.99 g/mol. \(m = n \cdot \text{MM} = (0.50\,\text{mol})(105.99\,\text{g/mol}) = 52.995\,\text{g}\)

Dissolve the salt in water

Weigh 52.995 g of sodium carbonate and dissolve it in enough water to reach a final volume of 1.00 L.

Key concepts

Dilution Formula

When preparing solutions from concentrated reagents, the dilution formula is an essential tool. It allows us to calculate how much of the concentrated form we need in order to achieve our desired concentration and volume. The dilution formula is:\[ M_1V_1 = M_2V_2 \]In this equation, \( M_1 \) and \( V_1 \) correspond to the molarity and volume of the concentrated solution, while \( M_2 \) and \( V_2 \) are the desired molarity and final volume after dilution. The equation assumes that the amount of solute remains constant before and after dilution. This is particularly useful for strong acids and bases, which are often stored in concentrated forms for efficiency.- The process involves: - Identifying initial and final concentrations. - Solving for the unknown volume or concentration. - Carefully adding the concentrated solution to a measured quantity of water until the final volume is reached.
Precision is important. Using too much or too little of the concentrated solution will alter the final concentration. Always ensure that calculations are double-checked before proceeding.

Molarity Calculations

Molarity is a measure of concentration expressing the number of moles of solute per liter of solution. It is denoted by \( M \) and is calculated using the formula:\[ M = \frac{n}{V} \]Here, \( n \) represents the number of moles of solute, and \( V \) is the volume of the solution in liters. Molarity is a commonly used concentration unit in chemistry because it provides a straightforward way to relate solute to solution.- Steps to calculate molarity involve: - Determining the number of moles, typically through measured mass or given volume. - Accurately measuring the solution's total volume.A high molarity indicates a concentrated solution, while a low molarity indicates a dilute solution. Understanding the balance between concentration and volume is critical in preparing accurate chemical solutions.

Molar Mass

Molar mass is the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. For example, the molar mass of sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) involves adding the atomic masses of sodium, carbon, and oxygen from the periodic table.- To find molar mass: - Consider all constituent atoms of the compound. - Use atomic weights from a reliable periodic table.Molar mass is crucial when converting between moles and grams, which is often required in chemistry calculations. For solid reagents, knowing the molar mass ensures accurate measurement and use in molarity calculations.

Concentration

Concentration tells us how much solute is present in a given amount of solution. It provides essential information for reacting solutions or comparing their strengths. While molarity is a common unit, concentration can also be expressed as mass/volume percentage, molality (moles per kilogram of solvent), and more. - Importance of concentration: - Determines solution strength and potential reaction outcomes. - Influences the direction and rate of chemical reactions. Concentration's impact on chemical properties is immense. For example, solutions with higher concentrations are often more reactive and can be hazardous. In laboratory settings, precise concentration adjustments are vital for safety and accuracy.

Dissolving Solids

Creating a solution from a solid involves dissolving the correct amount into a solvent, usually water. This process requires knowledge of the desired concentration and the solute's molar mass.- Steps to dissolve solids: - Calculate the required mass using the formula: mass = moles \( \times \) molar mass. - Weigh out the exact amount of solid using a precision balance. - Add the solid to a specific volume of solvent, ensuring thorough mixing until dissolved.The goal is to achieve uniform distribution of solute particles throughout the solvent, resulting in a homogeneous solution. Factors like solubility, temperature, and stirring can affect how well and quickly a solute dissolves. Proper technique here is key to creating solutions with the intended properties.