Problem 117

Question

How many milliliters of concentrated hydrochloric acid solution ( \(36.0 \% \mathrm{HCl}\) by mass, density \(=1.18 \mathrm{~g} / \mathrm{mL}\) ) are required to produce \(10.0 \mathrm{~L}\) of a solution that has a \(\mathrm{pH}\) of \(2.05\) ?

Step-by-Step Solution

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Answer
The required volume of concentrated HCl solution needed to make the 10 L solution with a pH of 2.05 is 7.77 mL.
1Step 1: 1: Find the concentration of hydronium ions
Using the given pH = 2.05, calculate the concentration of hydronium ions \([H_3O^{+}]\) in the final solution using the relationship: \(pH = -\log[H_3O^{+}]\) \[2.05 = -\log[H_3O^{+}]\] To find \([H_3O^{+}]\), we can rewrite the equation as: \([H_3O^{+}] = 10^{-pH}\) \([H_3O^{+}] = 10^{-2.05}\] \([H_3O^{+}] = 8.91\times 10^{-3}\mathrm{~M}\)
2Step 2: 2: Calculate moles of HCl
To find the moles of HCl required to produce the 10.0 L solution with the concentration of hydronium ions \([H_3O^{+}]\) found in step 1, use the formula: \(moles = [H_3O^{+}] \times volume\) \(moles = (8.91\times 10^{-3}\mathrm{~M}) (10.0\mathrm{~L})\) \(moles = 0.0891\mathrm{~mol}\)
3Step 3: 3: Determine the mass of HCl
To find the mass of HCl required, we will use the molar mass of HCl = 36.46 g/mol and the moles found in step 2: \(mass = moles \times molar\,mass\) \(mass = (0.0891\mathrm{~mol})(36.46 \mathrm~{g/mol})\) \(mass = 3.25 \mathrm{~g}\)
4Step 4: 4: Calculate the volume of concentrated HCl solution needed
We are given that the concentrated HCl solution is 36.0% HCl by mass, and has a density of 1.18 g/mL. To find the volume of the concentrated HCl solution needed to obtain the calculated mass of HCl in step 3, use the relationship: \(volume = \frac{mass}{(fraction\_of\_\%HCl) (density)}\) \(volume = \frac{3.25 \mathrm{~g}}{(0.36)(1.18 \mathrm{~g/mL})}\) \(volume = 7.77 \mathrm{~mL}\) The required volume of concentrated HCl solution needed to make the 10 L solution with a pH of 2.05 is 7.77 mL.

Key Concepts

Hydronium Ion ConcentrationMole CalculationHydrochloric Acid Solution
Hydronium Ion Concentration
Understanding the concept of hydronium ion concentration is key to grasping how acidic or basic a solution is. The hydronium ion, represented as \([H_3O^+]\), directly influences the pH level of solutions. A higher concentration of these ions results in a lower pH, indicating a more acidic environment.
To calculate the hydronium ion concentration given a pH, the formula is:
  • \( [H_3O^+] = 10^{-pH} \)
For a solution with a pH of 2.05, the concentration can be calculated as \( 8.91 \times 10^{-3} \text{ M} \). This means there are \( 8.91 \times 10^{-3} \) moles of hydronium ions per liter of solution.
Knowing this offers insight into how much of an acid or base needs to be added to alter the pH of the solution. It’s crucial for tasks involving chemistry experiments or solving real-world chemistry problems like balancing acid rain.
Mole Calculation
Mole calculation is a fundamental concept in chemistry that allows us to quantify atoms, ions, or molecules in a given sample. When dealing with solutions and reactions, understanding how to calculate moles is vital.
The basic formula used for calculating the number of moles is:
  • \( \text{moles} = \text{concentration of solute (M)} \times \text{volume of solution (L)} \)
In the context of the exercise, knowing the concentration of hydronium ions as \( 8.91 \times 10^{-3} \text{ M} \), and the volume of the solution as 10 liters, the moles of HCl required are \( 0.0891 \text{ mol} \).
This step involves multiplying the molarity (M) with the volume in liters to get the total number of moles.
Using this principle helps you calculate not only in scenarios of dilution or solution preparation but is also useful in stoichiometry for balanced chemical reactions.
Hydrochloric Acid Solution
Hydrochloric acid (HCl) is a strong acid used in various applications from cleaning to scientific research. In solutions, its behavior depends significantly on its concentration and the way it is prepared.
When you have a concentrated hydrochloric acid solution, which might be labeled by mass percentage (such as 36% \( \text{%} \) HCl), you can determine how much of the solution is needed to reach a desired concentration in a diluted mixture.
The steps involve:
  • Calculating the needed mass of HCl using its molar mass (e.g., for the exercise, 3.25 g is needed).
  • Using the percentage by mass and density to find the volume of the concentrated solution required.
For instance, with a density of 1.18 g/mL and 36% concentration, you’d find the required volume as 7.77 mL to achieve the target acidity. This calculation is fundamental when adjusting solutions for experimental tasks, analysis, or even industrial processes.
Understanding these concepts assists in safe and effective usage and preparation of chemical solutions for any purpose.