1\. A 1.50 mL aliquot of a \(0.177 \mathrm{M}\) solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) is diluted into \(10.0 \mathrm{~mL}\) of distilled water, to give solution A. A \(10.0 \mathrm{~mL}\) aliquot of \(\mathbf{A}\) is then diluted into \(50.0 \mathrm{~mL}\) of distilled water, to give solution \(\mathbf{B}\). Finally, \(10.0 \mathrm{~mL}\) of \(\mathbf{B}\) is diluted into \(900.0 \mathrm{~mL}\) of distilled water to give solution C. Additional distilled water is then added to \(\mathrm{C}\) to give a final volume of \(1.0000 \mathrm{~L}\). What is the final concentration of sulfuric acid in solution C? 2\. A solution was prepared by mixing \(250 \mathrm{~mL}\) of \(0.547 \mathrm{M} \mathrm{NaOH}\) with \(50.0 \mathrm{~mL}\) of \(1.62 \mathrm{M} \mathrm{NaOH}\) and then diluting to a final volume of \(1.50 \mathrm{~L}\). What is the molarity of \(\mathrm{Na}^{+}\) in this solution? To what final volume should \(75.00 \mathrm{~mL}\) of \(0.889\) M HCl(aq) be diluted to prepare \(0.800 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) ?

Step by step solution

01

- Calculate the number of moles of sulfuric acid in the initial aliquot

To find the number of moles of H2SO4 in the initial 1.50 mL aliquot, use the molarity and volume of the solution:\[ \text{moles of H2SO4} = \text{Molarity (M)} \times \text{Volume (L)} \]. Convert the volume from mL to L by dividing by 1000:\[ \text{Volume} = \frac{1.50\,\text{mL}}{1000} = 0.0015\,\text{L} \]. Calculate the moles:\[ \text{moles of H2SO4} = 0.177\,M \times 0.0015\,L = 0.0002655\,\text{moles} \].

02

- Apply the dilution concept for solution A

Remember that the number of moles of solute remains the same upon dilution. The diluted concentration in solution A after adding to 10.0 mL of water can be overlooked, since we are concerned with the moles of solute.

03

- Calculate the moles in solution B

After taking 10.0 mL from solution A and diluting in 50.0 mL of water, the same moles of H2SO4 from step 1 are present in a new volume of (10.0 + 50.0) mL or 0.060 L.

04

- Calculate the moles in solution C

Diluting 10.0 mL of solution B into 900.0 mL of water still retains the moles of solute. The new volume is 10.0 mL + 900.0 mL = 910.0 mL or 0.910 L. Additional water added to reach 1.000 L does not change the number of moles but increases volume to 1.000 L.

05

- Calculate the final concentration of sulfuric acid in solution C

Using the initial moles calculated and the final volume, calculate the final concentration:\[ \text{Concentration of H2SO4 in C} = \frac{\text{moles of H2SO4}}{\text{Volume of solution C (L)}} = \frac{0.0002655\,\text{moles}}{1.000\,L} = 0.0002655\,M \].

06

- Find the total moles of NaOH from both solutions before mixing

First, calculate the moles in each NaOH solution separately:\[ \text{moles in 250 mL of 0.547 M NaOH} = 0.547\,M \times \frac{250\,mL}{1000} = 0.13675\,\text{moles} \] and \[ \text{moles in 50.0 mL of 1.62 M NaOH} = 1.62\,M \times \frac{50.0\,mL}{1000} = 0.081\,\text{moles} \]. Then add the moles together.

07

- Calculate the total volume after mixing NaOH solutions

The total volume after mixing and diluting is the final volume of 1.50 L.

08

- Calculate the molarity of Na+ in the final NaOH solution

Add up the moles of NaOH from the two solutions and divide by the final volume to find the molarity of the NaOH solution. Since each NaOH contributes one Na+, this will also be the molarity of Na+.\[ \text{Total moles of NaOH} = 0.13675\,\text{moles} + 0.081\,\text{moles} = 0.21775\,\text{moles} \] \[ \text{Molarity of Na+} = \frac{\text{Total moles of NaOH}}{\text{Volume (L)}} = \frac{0.21775\,\text{moles}}{1.50\,L} = 0.14517\,M \].

09

- Calculate the volume to dilute HCl to get 0.800 M concentration

For the last part, use the dilution equation \( M_1V_1 = M_2V_2 \) where \(M_1\) and \(V_1\) are the molarity and volume of the concentrated solution and \(M_2\) and \(V_2\) are the molarity and volume of the diluted solution. Solve for \(V_2\) to find the required final volume for the HCl solution.\[ V_2 = \frac{M_1V_1}{M_2} = \frac{0.889\,M \times 75.00\,mL}{0.800\,M} = \frac{66.675}{0.800} = 83.34375\,mL\].

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Key Concepts

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

Molarity

Molarity is a fundamental concept in chemistry that defines the concentration of a solution. It is the number of moles of solute (the substance being dissolved) per liter of solution. The molarity (M) is mathematically expressed as:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
For example, if a solution has a molarity of 0.177 M, as given in the exercise for sulfuric acid, it means that there are 0.177 moles of sulfuric acid in every liter of solution.

Importance in Calculations

Molarity simplifies quantifying the concentration of solutions and allows for straightforward calculations during dilutions and mixing, hence its ubiquity in laboratory settings. When diluting or mixing solutions, one can easily calculate the resulting molarity and hence understand the strength of the resulting solution.

Dilution of Solutions

The process of dilution involves reducing the concentration of a solute in a solution by adding more solvent. While the amount of solute remains constant, the total volume of the solution increases, thus lowering the molarity. The relationship between the initial and final concentrations and volumes is given by the dilution equation:
\[ M_1V_1 = M_2V_2 \]
Where \( M_1 \) and \( V_1 \) represent the initial molarity and volume, and \( M_2 \) and \( V_2 \) denote the final molarity and volume respectively. This formula shows that the product of the initial concentration and volume equals the product of the final concentration and volume.

Practical Example

When Solution A is diluted to form Solution B and then Solution C in the exercise, the overall moles of solute remain unchanged, but the volume of solvent increases. Knowing this principle helps us understand that diluting a concentrated sulfuric acid into a larger volume of water will not affect the total moles of sulfuric acid present.

Moles of Solute

In chemistry, the mole is a unit that quantifies the amount of a substance. It's one of the seven base SI units and is defined as the amount of any chemical substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 grams of carbon-12.
The number of moles of a solute in a solution plays a pivotal role in determining the solution's concentration. The step-by-step solution starts by calculating the moles of sulfuric acid from a given molarity and volume. This calculation is crucial for any further calculation involving the concentration of this solute in various dilution stages.'

Calculation Example

Using the formula \( moles = molarity \times volume \), we can determine the number of moles of solute present in any volume of the solution. For instance, to determine the initial moles of sulfuric acid in the exercise, you would multiply the molarity (0.177 M) by the volume in liters (0.0015 L), giving you 0.0002655 moles.

Concentration Calculation

Concentration calculation is a critical exercise in chemistry that allows scientists to understand the makeup of a solution. The concentration can be expressed in various ways, with molarity being one of the most common. Knowing the concentration helps in predicting the behavior of a substance in a reaction and for preparing solutions of desired strength.
To calculate the concentration of a dilute solution, as observed in Solution C for sulfuric acid, one must use the number of moles of the substance and the total volume of the solution after dilution. Once you have these values, the concentration calculation becomes straightforward, using the molarity formula provided earlier.

• Identify the moles of solute initially present.
• Understand that dilution does not change the moles of solute.
• Calculate the total volume of the solution after dilution.
• Divide the moles of solute by the total volume to find the new concentration.

By understanding these steps, students can navigate through various dilution problems and accurately arrive at the final concentrations of diluted solutions.