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Suppose you have \(557 \mathrm{~mL}\) of \(0.0300 \mathrm{M} \mathrm{HCl}\), and you want to make up a solution of \(\mathrm{HCl}\) that has a \(\mathrm{pH}\) of 1.751. What is the maximum volume (in liters) that you can make of this solution?

Short Answer

Expert verified
The maximum volume of the solution you can prepare is approximately 0.939 L.

Step by step solution

01

Calculate Target Hydrogen Ion Concentration

To find the concentration of hydrogen ions for the target \(\text{pH}\) of 1.751, use the formula \(\text{pH} = -\log[\text{H}^+]\). Rearrange it to find \([\text{H}^+]\): \([\text{H}^+] = 10^{-\text{pH}}\). Plug in the \(\text{pH}\) value: \([\text{H}^+] = 10^{-1.751}\).
02

Compute \([\text{H}^+]\) for the Desired pH

Calculating \([\text{H}^+]\), we find that \([\text{H}^+] = 10^{-1.751} \approx 0.0178 \space \text{M}\). This is the concentration of \(\text{HCl}\) needed in the final solution.
03

Set Up Dilution Equation

Use the dilution equation \( C_1V_1 = C_2V_2 \) where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume. Here, \(C_1 = 0.0300 \space \text{M}\), \(V_1 = 0.557 \space \text{L}\), and \(C_2 = 0.0178 \space \text{M}\). We solve for \(V_2\).
04

Calculate Maximum Volume

Rearrange the dilution equation to solve for \(V_2\): \(V_2 = \frac{C_1V_1}{C_2}\). Substitute the known values: \(V_2 = \frac{0.0300 \times 0.557}{0.0178}\). Calculate \(V_2\) to find the final volume in liters.
05

Compute and Conclude

Carrying out the calculation: \(V_2 = \frac{0.01671}{0.0178} \approx 0.939 \space \text{L}\). This is the maximum volume of the solution with a \(\text{pH}\) of 1.751 that you can prepare using the available \(0.0300 \space \text{M} \space \text{HCl}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

HCl concentration
Hydrochloric acid (HCl) concentration refers to the amount of HCl present in a given volume of solution. Concentration is typically expressed in molarity (M), which is moles of solute per liter of solution. For example, if you have a 0.0300 M HCl solution, it means that there are 0.0300 moles of HCl in every liter of the solution.
Understanding concentration is crucial in chemistry as it helps in determining how much reactant you have during a chemical reaction. In a laboratory setting, knowing the concentration allows precise interaction with other chemicals and accurate preparation of desired solutions.
For instance, in the given exercise, our starting point is a 0.0300 M HCl solution, which will be manipulated to reach a desired pH value of 1.751.
pH calculations
The pH of a solution is a measure of its acidity or basicity, on a scale from 0 to 14. The lower the pH, the higher the acidity. To determine the pH, chemists use the formula:
  • \[ ext{pH} = - ext{log}[ ext{H}^+] \]
where \([ ext{H}^+]\) is the concentration of hydrogen ions.
In the context of making a solution with a desired pH, it's necessary to compute the hydrogen ion concentration that corresponds to that pH. For example, with a target pH of 1.751, we calculate:
  • \([ ext{H}^+] = 10^{-1.751} \approx 0.0178 ext{ M}\)
This informs us about the required acidic strength of the solution we aim to prepare.
hydrogen ion concentration
The concentration of hydrogen ions \([ ext{H}^+]\) is key in defining the acidity of a solution. A higher hydrogen ion concentration signifies a more acidic solution and corresponds to a lower pH. The relationship between pH and hydrogen ion concentration is exponential, as seen in the equation:
  • \([ ext{H}^+] = 10^{- ext{pH}}\)
In our exercise, to achieve a pH of 1.751, the calculated hydrogen ion concentration is 0.0178 M. This determines the HCl concentration we require in the solution.
It's important to understand this concept as it directly affects solution preparation, influencing the dilution and adjustment of the initial HCl concentration.
solution preparation
Solution preparation is a fundamental skill in chemistry that involves calculating and mixing the right amounts of substances to achieve a desired concentration, volume, or pH. It often involves using the dilution equation \( C_1V_1 = C_2V_2 \) to determine how to adjust an existing solution.
In this equation:
  • \(C_1\) is the initial concentration,
  • \(V_1\) is the initial volume,
  • \(C_2\) is the desired concentration,
  • \(V_2\) is the final volume you can prepare.
For example, with an initial solution of 0.0300 M HCl at a volume of 0.557 L, our goal is to prepare a new solution with 0.0178 M concentration (for a pH of 1.751).
Rearranging the dilution equation to solve for \(V_2\) gives:
  • \[ V_2 = \frac{C_1V_1}{C_2} \]
By plugging in the given values, the calculation results in a new solution volume of approximately 0.939 L.
This tells us how much solution can be made while achieving the desired acidity level.

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Most popular questions from this chapter

You want to make up \(3.00 \mathrm{~L}\) of aqueous hydrochloric acid, \(\mathrm{HCl}(a q),\) that has a \(\mathrm{pH}\) of \(2.00 .\) How many grams of concentrated hydrochloric acid will you need? Concentrated hydrochloric acid contains 37.2 mass percent of \(\mathrm{HCl}\).

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