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What yolume of \(0.25 M \mathrm{HCl}\) solution must be diluted to prepare \(1.00 \mathrm{~L}\) of \(0.040 \mathrm{M} \mathrm{HCl}\) ?

Short Answer

Expert verified
To prepare 1.00 L of 0.040 M HCl, 0.16 L of 0.25 M HCl solution must be diluted.

Step by step solution

01

Write the dilution formula

Write down the formula for dilution: \(C_1V_1 = C_2V_2\)
02

Substitute the given values

Plug in the given values of initial and final concentrations and the final volume of the diluted solution: \((0.25 \mathrm{M})V_1 = (0.040 \mathrm{M})(1.00 \mathrm{L})\)
03

Solve for the initial volume \(V_1\)

To find the initial volume of the HCl solution (\(V_1\)), divide both sides by the initial concentration (0.25 M): \(V_1 = \frac{(0.040 \mathrm{M})(1.00 \mathrm{L})}{(0.25 \mathrm{M})}\)
04

Calculate the result

Now, simplify and calculate the value of \(V_1\): \(V_1 = \frac{(0.040 \mathrm{M})(1.00 \mathrm{L})}{(0.25 \mathrm{M})} = 0.16 \mathrm{L}\)
05

Express the result

Finally, we express the result in a clear manner: To prepare 1.00 L of 0.040 M HCl, 0.16 L of 0.25 M HCl solution must be diluted.

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Key Concepts

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

Understanding Molarity
Molarity is a measure of concentration used in chemistry to express the amount of a solute dissolved in a solution. Imagine molarity as a way to describe how strong or weak a solution is. It is represented by the letter "M" and defined as the number of moles of solute per liter of solution. The formula to calculate molarity is:
\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]This measurement helps chemists to understand and convey the "strength" of a chemical solution. For example, a solution labeled as \(0.40 \text{M HCl}\) contains 0.40 moles of hydrochloric acid in every liter of the solution.
Molarity is often needed when preparing solutions in a laboratory setting. When diluting a solution, you aim for a specific molarity to achieve desired chemical reactions or properties. This process involves adjusting the amount of solute and solvent to reach the needed concentration.
Exploring Concentration
Concentration refers to how much of a substance (the solute) is present in a given amount of space (the solvent or solution). It's like asking, "How salty is your soup?" The more salt, the higher the concentration.
In a laboratory, concentration is crucial because it affects reaction rates, chemical balance, and safety. Solutions with different concentrations can have vastly different properties and uses. For example, a concentrated hydrochloric acid solution is much more corrosive and reactive than a diluted one.
  • Concentration can be expressed in different ways, such as molarity, as mentioned above, or percentage by mass.
  • The ability to calculate and adjust concentrations is vital for performing dilutions and mixture preparations safely and accurately.
When diluting, you're increasing the solvent amount to decrease the solute concentration without changing the solute amount. This is crucial when specific solution strength is needed for experiments or practical applications.
Volume Calculations in Dilution
Volume calculations are a key step in performing a dilution. These calculations help determine the amount of solution needed to reach a desired concentration. In the dilution formula, \(C_1V_1 = C_2V_2\), each symbol has a specific meaning:
  • \(C_1\) is the initial concentration of the solution.
  • \(V_1\) is the initial volume of the solution.
  • \(C_2\) is the concentration after dilution.
  • \(V_2\) is the final volume after dilution.
To solve for an unknown, rearrange the formula to isolate the variable of interest. In dilution problems, you often know most of these values and need to calculate one unknown. For instance, to find the initial volume \(V_1\) needed to prepare a less concentrated solution, you'd use:
\[V_1 = \frac{C_2V_2}{C_1}\]By plugging in the known values, you can solve for the unknown. Always double-check that your units are correct and remember that in dilution problems, adjusting volumes is key to changing concentrations without altering the amount of solute.

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Most popular questions from this chapter

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Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C}\) ? a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}_{3} \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol [ 143 torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water. \()\)

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