Question

Many laboratories keep bottles of \(3.0 \mathrm{M}\) solutions of the common acids on hand. Given the following molarities of the concentrated acids, determine how many milliliters of each concentrated acid would be required to prepare \(225 \mathrm{mL}\) of a \(3.0 \mathrm{M}\) solution of the acid. $$ \text {Acid} \quad \quad \text {Molarity of Concentrated Reagent} $$ $$ \text {\(\mathrm{HCl}\)} \quad \quad \text {\(12.1 M\)} $$ $$ \text {\(\mathrm{HNO}_{3}\) } \quad \quad \text {\(15.9 M\)} $$ $$ \text {\(\mathrm{H}_{2} \mathrm{SO}_{4}\)} \quad \quad \text {\(18.0 M\)} $$ $$ \text {\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)} \quad \quad \text {\(17.5 M\)} $$ $$ \text {\(\mathrm{H}_{3} \mathrm{PO}_{4}\) } \quad \quad \text {\(14.9 M\)} $$

Short answer

The required volumes for each concentrated acid to prepare 225 mL of a 3.0 M solution are: HCl: \( \approx 55.79\,\mathrm{mL} \), HNO3: \( \approx 42.52\,\mathrm{mL} \), H2SO4: \( \approx 37.50\,\mathrm{mL} \), HC2H3O2: \( \approx 38.57\,\mathrm{mL} \), and H3PO4: \( \approx 45.30\,\mathrm{mL} \).

Step-by-step solution

Find volume for HCl

To find the milliliters of 12.1 M HCl required, use the formula: \( C_1V_1 = C_2V_2\) Plug in the values: \( (12.1\,\mathrm{M})V_1 = (3.0\,\mathrm{M})(225\,\mathrm{mL}) \) Now, solve for \(V_1\): \( V_1 = \frac{(3.0\,\mathrm{M})(225\,\mathrm{mL})}{(12.1\,\mathrm{M})} \) \( V_1 \approx 55.79\,\mathrm{mL} \)

Find volume for HNO3

To find the milliliters of 15.9 M HNO3 required, use the formula: \( C_1V_1 = C_2V_2\) Plug in the values: \( (15.9\,\mathrm{M})V_1 = (3.0\,\mathrm{M})(225\,\mathrm{mL}) \) Now, solve for \(V_1\): \( V_1 = \frac{(3.0\,\mathrm{M})(225\,\mathrm{mL})}{(15.9\,\mathrm{M})} \) \( V_1 \approx 42.52\,\mathrm{mL} \)

Find volume for H2SO4

To find the milliliters of 18.0 M H2SO4 required, use the formula: \( C_1V_1 = C_2V_2\) Plug in the values: \( (18.0\,\mathrm{M})V_1 = (3.0\,\mathrm{M})(225\,\mathrm{mL}) \) Now, solve for \(V_1\): \( V_1 = \frac{(3.0\,\mathrm{M})(225\,\mathrm{mL})}{(18.0\,\mathrm{M})} \) \( V_1 \approx 37.50\,\mathrm{mL} \)

Find volume for HC2H3O2

To find the milliliters of 17.5 M HC2H3O2 required, use the formula: \( C_1V_1 = C_2V_2\) Plug in the values: \( (17.5\,\mathrm{M})V_1 = (3.0\,\mathrm{M})(225\,\mathrm{mL}) \) Now, solve for \(V_1\): \( V_1 = \frac{(3.0\,\mathrm{M})(225\,\mathrm{mL})}{(17.5\,\mathrm{M})} \) \( V_1 \approx 38.57\,\mathrm{mL} \)

Find volume for H3PO4

To find the milliliters of 14.9 M H3PO4 required, use the formula: \( C_1V_1 = C_2V_2\) Plug in the values: \( (14.9\,\mathrm{M})V_1 = (3.0\,\mathrm{M})(225\,\mathrm{mL}) \) Now, solve for \(V_1\): \( V_1 = \frac{(3.0\,\mathrm{M})(225\,\mathrm{mL})}{(14.9\,\mathrm{M})} \) \( V_1 \approx 45.30\,\mathrm{mL} \) Hence, the required volumes for each concentrated acid are as follows: HCl: \( \approx 55.79\,\mathrm{mL} \) HNO3: \( \approx 42.52\,\mathrm{mL} \) H2SO4: \( \approx 37.50\,\mathrm{mL} \) HC2H3O2: \( \approx 38.57\,\mathrm{mL} \) H3PO4: \( \approx 45.30\,\mathrm{mL} \)

Key concepts

Solution Preparation

Preparing solutions with precise concentrations is a foundational task in chemistry, as it ensures that experiments and reactions can be performed consistently. A solution is made up of a solute, which is the substance to be dissolved, and a solvent, the medium in which the solute is dispersed. When preparing a solution, one must consider the desired molarity, volume, and the concentration of the starting material.

For example, when a laboratory needs a specific molarity of acid for a reaction, technicians use concentrated reagents—high-molarity solutions of the acids. They must calculate the exact amount of this concentrated acid to dilute with a solvent, often water, to achieve the final desired concentration. The calculations shown previously illustrate how to determine the volume of concentrated acid needed to make a new solution with a specified volume and molarity. Always remember to add the acid to water, not the other way around, to prevent exothermic reactions that may cause splashing or boiling.

Concentration of Solutions

The concentration of a solution describes how much solute is present in a given volume of solvent. There are various ways to express concentration, but molarity is one of the most common, especially in the context of aqueous solutions. Molarity is defined as moles of solute per liter of solution, with the unit of measurement being moles per liter (M).

Understanding solution concentration is vital, as it affects the rate, extent, and safety of chemical reactions. When precise measurements are necessary, it is important to properly calculate and adjust concentrations. This application of molarity in calculating volumes for dilution guides many laboratory procedures, from routine experiments to the formulation of medicines.

Dilution of Acids

Dilution is the process of reducing the concentration of a solution by adding more solvent. With acids, dilution should be carried out carefully to avoid heat and splashes caused by the exothermic dissolving process. To dilute an acid, we must use the equation \( C_1V_1 = C_2V_2 \) where \( C_1 \) and \( C_2 \) represent the initial and final concentrations, while \( V_1 \) and \( V_2 \) stand for the initial and final volumes, respectively.

This formula helps calculate how much of a concentrated stock solution (like the concentrated acids mentioned in the exercise) is needed to make a new diluted solution with a specific volume and molarity. Once \( V_1 \) is found using the equation, one can proceed with the careful addition of the concentrated acid into the appropriate volume of water to achieve the targeted dilution.

Molar Concentration

Molar concentration, or molarity, represents the number of moles of a substance per liter of solution. It provides an easy way to relate amounts of reactants and products in a chemical reaction. When we know the molarity of a solution, we can use it to calculate how many grams of a substance are in a given volume of solution, or how much volume is needed to have a certain number of moles of substance.

In the textbook solutions, we utilized molarity to find out how much of a concentrated acid is required to make a solution with a lower, specific molarity. This involves rearranging the molarity equation for the unknown volume. For instance, steps are shown for calculating the needed volume of concentrated hydrochloric acid (HCl) to prepare a new solution with a molarity of 3.0 M. This kind of molarity calculation is routine in laboratory settings and is a key skill in solution preparation and analysis.