Problem 72 - General Chemistry

Question

How many milliliters of \(0.126 \mathrm{M} \mathrm{HClO}_{4}\) (perchloric acid) are required to give \(0.102 \mathrm{~mol} \mathrm{HClO}_{4}\) ?

Step-by-Step Solution

Answer

810 mL of \(0.126 \mathrm{M} \mathrm{HClO}_{4}\) are needed.

Step 1: Understand the Relationship

We are given a molarity and a number of moles and need to find the volume. Molarity \(M\) is the number of moles of solute per liter of solution, expressed as \(M = \frac{n}{V}\), where \(n\) is the number of moles and \(V\) is the volume in liters.

Step 2: Rearrange the Formula

To find the volume \(V\), rearrange the molarity formula: \( V = \frac{n}{M} \). This will allow us to calculate the volume needed for the given number of moles and molarity.

Step 3: Plug in Known Values

Substitute the given values into the equation: the number of moles, \(n = 0.102\) mol, and molarity, \(M = 0.126\) M. The equation becomes \( V = \frac{0.102}{0.126} \).

Step 4: Calculate the Volume in Liters

Compute the volume in liters using the values: \( V = \frac{0.102}{0.126} = 0.8095238... \) liters. Round the answer to a reasonable number of significant figures, which based on the values given is \(0.810\) liters.

Step 5: Convert Liters to Milliliters

Since 1 liter equals 1000 milliliters, multiply the volume in liters by 1000 to convert to milliliters: \(0.810 \times 1000 = 810\) mL.

Key Concepts

Calculating volume from molarity

When it comes to calculating the volume from molarity, the key formula to use is based on the relationship between molarity, moles of solute, and volume of solution. Molarity (M) is defined as the number of moles of a solute per liter of solution. The formula is:\[ M = \frac{n}{V} \]where:

• \( M \) is the molarity of the solution,
• \( n \) is the number of moles of the solute, and
• \( V \) is the volume of the solution in liters.

To find the volume, you rearrange this formula to solve for \( V \):\[ V = \frac{n}{M} \]This allows us to calculate the volume needed if we know the number of moles of solute and the molarity of the solution. It is important to first ensure that the number of moles is correctly calculated and then to apply the molarity correctly in this formula. After computing the volume in liters, remember to convert to milliliters if needed by multiplying by 1000, since 1 liter is equivalent to 1000 milliliters.

Perchloric acid

Perchloric acid, chemically known as \( HClO_4 \), is a strong acid commonly encountered in labs and used in various chemical reactions. It is colorless and, in its concentrated form, is a highly corrosive and dangerously reactive liquid. Despite its hazards, it is a valuable reagent in measurements and reactions due to its strong acidic nature.

Here are some key facts about perchloric acid:

• Often used in etching and cleaning in the electronics industry.
• Acts as a catalyst in different chemical reactions.
• When dilute, perchloric acid behaves similarly to other strong acids like sulfuric and nitric acids.

Given its high reactivity, special care must be taken when handling \( HClO_4 \) to avoid chemical accidents and ensure lab safety.

Stoichiometry in solution chemistry

Stoichiometry deals with the calculation of reactants and products in chemical reactions. In the realm of solution chemistry, stoichiometry helps us understand the quantitative relationships in chemical reactions involving solutions. It requires the use of molarity to convert between volumes and moles, thus allowing for precise calculations.

When performing stoichiometric calculations:

• First, identify the balanced chemical equation for the reaction.
• Determine the molarity and volume of the reactants involved.
• Convert volumes to moles using the molarity, then use stoichiometric ratios from the balanced equation to find moles of other substances involved.
• Finally, if necessary, convert moles back to a desired unit, such as mass or volume.

This structured approach ensures that all aspects of a reaction are considered and provides an accurate picture of the quantities involved in each step of the process.