Question

How many moles of \(\mathrm{H}^{+}\) ions are present in each of the following aqueous solutions? (a) \(1.40 \mathrm{~L}\) of \(0.25 \mathrm{M}\) perchloric acid (b) \(6.8 \mathrm{~mL}\) of \(0.92 \mathrm{M}\) nitric acid (c) \(2.6 \mathrm{~L}\) of \(0.085 \mathrm{M}\) hydrochloric acid

Short answer

a) 0.35 moles, b) 0.006256 moles, c) 0.221 moles.

Step-by-step solution

Use the formula to find moles of \(\mathrm{H}^{+}\) ions

The concentration (M) provides the moles of solute per liter. The formula to find moles is: \[\text{{Moles of }} \mathrm{H}^{+} = \text{{Volume (in liters)}} \times \text{{Concentration (M)}} \]

Calculate moles for part (a)

Given: \(1.40 \mathrm{~L} \text{{ of }} 0.25 \mathrm{M}\text{{ perchloric acid}}\), use the formula: \[1.40 \mathrm{~L} \times 0.25 \mathrm{M} = 0.35 \text{{ moles of }} \mathrm{H}^{+} \]

Convert mL to L for part (b)

Given: \(6.8 \mathrm{~mL} \text{{ of }} 0.92 \mathrm{M} \text{{ nitric acid}}\), convert mL to L: \[6.8 \mathrm{~mL} = 6.8 \times 10^{-3} \mathrm{~L} = 0.0068 \mathrm{~L} \]

Calculate moles for part (b)

Use the formula: \[0.0068 \mathrm{~L} \times 0.92 \mathrm{M} = 0.006256 \text{{ moles of }} \mathrm{H}^{+} \]

Calculate moles for part (c)

Given: \(2.6 \mathrm{~L} \text{{ of }} 0.085 \mathrm{M} \text{{ hydrochloric acid}}\), use the formula: \[2.6 \mathrm{~L} \times 0.085 \mathrm{M} = 0.221 \text{{ moles of }} \mathrm{H}^{+} \]

Key concepts

molarity

Molarity is a measure of the concentration of a solute in a solution and is expressed as moles of solute per liter of solution. It is denoted by the symbol **M**.
Molarity is essentially a ratio that represents the amount of a substance (the solute) dissolved in a certain volume of solution. The formula for molarity is:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
This means if you know the amount of solute in moles and the volume of the solution in liters, you can easily calculate the molarity.
In the given exercise, we use the molarity of the acid to find the number of moles of hydrogen ions \((\text{H}^+)\) in specific volumes of the solutions.
To find the number of moles from molarity and volume, we use the formula:
\[\text{Moles of } \text{H}^+ = \text{Volume (in liters)} \times \text{Concentration (M)} \].
Ensuring correct unit conversions (to liters) is essential for accurate calculations.

acid solutions

Acid solutions are characterized by their ability to donate protons \((\text{H}^+)\), and their strength can be described by their concentration.
In the context of the provided problems, various acids (perchloric acid, nitric acid, and hydrochloric acid) are given in different molar concentrations.
Acids dissociate in water to yield hydrogen ions \((\text{H}^+)\) and their corresponding anions. The strength of an acid in a solution relies on its ability to release \((\text{H}^+)\) ions.
For example:

• Perchloric acid (HClO4) dissociates completely in water to produce \((\text{H}^+)\) ions and \((\text{ClO}_4^-)\) ions. Given molarity tells us the concentration of \( \text{H}^+ \) in the solution.
• In nitric acid (HNO3), nitric acid molecules dissociate into \((\text{H}^+)\) ions and \((\text{NO}_3^-)\) ions.
• Hydrochloric acid (HCl) dissociates to yield \((\text{H}^+)\) and \((\text{Cl}^-)\) ions.

In each of these cases, the molarity of the acid directly gives the concentration of \( \text{H}^+ \) ions.

conversion of units

Converting units is a fundamental step in accurate calculative chemistry. In our exercise, we need to convert the volume from milliliters to liters for some parts of the problem.

To convert milliliters (mL) to liters (L), remember: there are 1000 milliliters in one liter.
So, the conversion factor is 1 liter/1000 milliliters, allowing us to write:
\[ V (\text{in liters}) = V (\text{in mL}) \times \frac{1}{1000} \]

For example, part (b) of the exercise involves:

Given: 6.8 mL of acid

Convert mL to L:
\[ 6.8 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0068 \text{ L} \]

Ensuring the correct units before performing molarity calculations is crucial for obtaining the right answer. Incorrect unit conversion can result in significant errors in your final result.