Question

What minimum volume of \(16 M\) sulfuric acid must be used to prepare \(750 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Short answer

The minimum volume of \(16\mathrm{M}\) sulfuric acid required to prepare \(750\mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is approximately \(4.69 \mathrm{~mL}\), calculated using the dilution formula \(C_{1}V_{1}=C_{2}V_{2}\).

Step-by-step solution

Write down the dilution formula

Dilution formula: \(C_{1}V_{1}=C_{2}V_{2}\).

Insert given values into the dilution formula

Substitute the values given for the concentrations and the desired volume: \(16 \mathrm{M} \times V_{1} = 0.1 \mathrm{M} \times 750 \mathrm{~mL}\).

Solve for the unknown volume

Now, you need to solve for \(V_{1}\), the volume of the concentrated acid: \[ V_{1} = \frac{0.1 \mathrm{M} \times 750 \mathrm{~mL}}{16 \mathrm{M}} \]

Calculate the volume and present the result

Calculate the volume of the concentrated acid required: \[ V_{1} = \frac{0.1 \mathrm{M} \times 750 \mathrm{~mL}}{16 \mathrm{M}} = 4.6875\mathrm{~mL} \] The minimum volume of \(16\mathrm{M}\) sulfuric acid required to prepare \(750\mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is approximately \(4.69 \mathrm{~mL}\).

Key concepts

Dilution Formula

Understanding the dilution formula is key to solving problems involving the preparation of solutions with a specific concentration. The formula is an expression of the conservation of mass, where the amount of solute remains unchanged during the dilition process. It's presented as:
\[ C_1V_1 = C_2V_2 \]
Here, \( C_1 \) and \( V_1 \) refer to the molarity and volume of the initial concentrated solution, while \( C_2 \) and \( V_2 \) correspond to the molarity and volume of the final diluted solution, respectively. One common application of this formula is in the laboratory when preparing solutions from stock concentrates, as shown in the given exercise. By rearranging the formula and solving for the unknown, students can find the volume or concentration needed to achieve the desired dilution.

Molarity

Molarity, symbolized as \( M \), is a measure of the concentration of a solution. It is defined as the moles of solute per liter of solution. This comes in handy when discussing solutions and their reactions. Molarity can be calculated using the formula:
\[M = \frac{moles~of~solute}{volume~of~solution~in~liters}\]
When working with molarity, keep in mind that it allows chemists to predict the outcome of reactions since it provides an exact measurement of how concentrated a solution is. It's crucial in understanding how solutions will behave and react with one another, especially when performing precise chemical experiments or producing specific concentrations of a solution.

Concentration of Solution

The concentration of a solution is a key concept that describes the amount of solute dissolved in a given quantity of solvent. It can be expressed in several ways, such as molarity, molality, normality, or simply as a percentage. The physical properties of solutions, like boiling point, osmotic pressure, and vapor pressure, are influenced by the concentration. In practical scenarios, quantifying concentration aids in ensuring the consistency and safety of products, from industrial applications to pharmaceutical formulations. It’s worth noting that concentration isn't fixed and can be adjusted by dilution or evaporation, thus altering the solution's properties.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the mole concept. Stoichiometry allows chemists to calculate the amounts of reactants needed or products formed in a chemical reaction. For instance, in the context of a solution preparation issue, stoichiometry can help delineate how much solute is required to achieve a desired molarity. This includes using the molar mass of the compound to convert between grams and moles, a fundamental step in many chemistry calculations. Understanding stoichiometry is also essential for grasping more advanced topics, like reaction yields and limiting reactants.