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Chapter 4: Problem 10

If you want to double the concentration of a solution, how could you do it? [Section 4.5\(]\)

Short Answer

Expert verified
To double the concentration of a solution, you can either reduce the volume of the solvent by half, double the amount of solute dissolved in it, or dilute a higher concentration solution with the appropriate volume of solvent. Calculation of the exact amount of solute or solvent needed can be done using the equation \(C_1V_1 = C_2V_2\).

Step by step solution

01

Scenario 1: Reducing the Volume of Solvent by Half

To double the concentration of the solution, the volume of the solvent can be reduced by half. This can be done by evaporating or removing half of the solvent. The solute will remain constant, hence resulting in a doubled concentration of the solution.
02

Scenario 2: Increasing the Amount of Solute

Another way to double the concentration of the solution is by doubling the amount of solute dissolved in it. The volume of the solvent will remain constant, but the overall concentration of the solution will be doubled as there is now twice the amount of solute.
03

Scenario 3: Diluting a Higher Concentration Solution

A third method to obtain a solution with double concentration can be achieved through diluting a higher concentration solution. This involves mixing an already prepared high concentration solution with an appropriate volume of solvent. The correct amount of high concentration solution must be calculated and then the solvent should be added accordingly to obtain a double concentrated solution.
04

Calculating the Required Amount of Solute or Solvent

In order to determine the exact amount of solute or solvent required to double the concentration of a solution, we can use the equation: \(C_1V_1 = C_2V_2\) where: - \(C_1\) = Initial concentration - \(V_1\) = Initial volume - \(C_2\) = Final concentration (twice the initial concentration) - \(V_2\) = Final volume By solving the equation for the desired variable (\(V_1\), \(V_2\), or the amount of solute), we can determine the exact amount of solute or solvent needed to double the concentration. For example, if we have a solution with a concentration of 0.1 M and a volume of 100 mL, we can plug the values into the equation to find out how much solvent should be removed to double the concentration: \(0.1 \times 100 = (0.1 \times 2) \times V_2\) \(V_2 = \frac{0.1 \times 100}{0.2} = 50\,mL\) So, in this case, we would need to remove 50 mL of solvent to double the concentration (or volume must be reduced to 50 mL).

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Key Concepts

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

Solvent Volume
When we talk about solvent volume, we refer to the amount of liquid in which the solute is dissolved. Changing the volume of the solvent impacts the concentration of the solution. To double the concentration of a solution, you can reduce the solvent volume by half. Imagine you have a lemonade and want it to taste twice as sweet, one option is to reduce the amount of water. Here are some ways to alter the solvent volume:
  • Evaporating the solvent: This method involves heating the solution so that the solvent turns into vapor, thereby reducing its volume.
  • Physically removing the solvent: This could be done by pouring out half of the liquid.
By reducing the volume, but keeping the total amount of solute the same, the concentration of the solution naturally increases. This is because concentration is defined as the amount of solute per unit volume of solvent.
Solute Amount
The solute amount is the quantity of substance that is dissolved in the solvent. To increase the concentration (or double it), another approach is to increase the amount of solute while keeping the solvent volume constant. For example, if you're making sugar water and want it to be sweeter, you simply add more sugar. Here鈥檚 how increasing the solute amount affects concentration:
  • Double the solute: By adding twice the initial amount of solute to the same volume of solvent, the concentration doubles.
  • Keep liquid volume constant: There's no need to change the solvent's volume; only the solute quantity needs adjusting.
It's important to ensure that the solute completely dissolves in the solvent for the concentration to be uniformly doubled throughout the solution.
Concentration Calculation
Concentration calculation is crucial for adjusting the amounts of solute and solvent to achieve a desired concentration. Calculating concentration involves using the formula: \(C_1 V_1 = C_2 V_2\). This equation stands for:
  • \(C_1\): Initial concentration of the solution.
  • \(V_1\): Initial volume of the solution.
  • \(C_2\): Desired final concentration, which in this context is double the initial.
  • \(V_2\): Final volume after the concentration adjustment.
For a practical example, if you start with a solution of concentration 0.1 M (moles per liter) and a volume of 100 mL, to double its concentration without changing the solute amount, you would solve:\[0.1 \times 100 = (0.1 \times 2) \times V_2\]\[V_2 = \frac{0.1 \times 100}{0.2} = 50 \, mL\]From this calculation, reducing the volume to 50 mL doubles the concentration. Understanding these calculations ensures precise adjustments to achieve the desired solution concentration.

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Most popular questions from this chapter

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathrm{b}) \mathrm{NO}_{3}^{-},(\mathrm{c}) \mathrm{NH}_{4}^{+},(\mathrm{d}) \mathrm{S}^{2-},\) (e) \(\mathrm{SO}_{4}^{2-}\). Explain briefly. [Section \(\left.4.2\right]\)

A 4.36 -g sample of an unknown alkali metal hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color signaling that the equivalence point has been reached after \(17.0 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the alkali metal cation: \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\) or \(\mathrm{Cs}^{+} ?\)

The arsenic in a 1.22-g sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in \(\mathrm{AsO}_{4}{ }^{3-} ?\) (b) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\)

Magnesium carbonate, magnesium oxide, and magnesium hydroxide are all white solids that react with acidic solutions. (a) Write a balanced molecular equation and a net ionic equation for the reaction that occurs when each substance reacts with a hydrochloric acid solution. (b) By observing the reactions in part (a) could you distinguish any of the three magnesium substances from the other two? If so how? (c) If excess \(\mathrm{HCl}(a q)\) is added, would the clear solutions left behind after the reaction is complete contain the same or different ions in each case?
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