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In the Laboratory Suppose you dilute \(25.0 \mathrm{mL}\) of a \(0.110 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to exactly \(100.0 \mathrm{mL}\). You then take exactly \(10.0 \mathrm{mL}\) of this diluted solution and add it to a \(250-\mathrm{mL}\) volumetric flask. After filling the volumetric flask to the mark with distilled water (indicating the volume of the new solution is \(250 .\) mL.), what is the concentration of the diluted \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution?

Short Answer

Expert verified
The concentration of the diluted solution is 0.00110 M.

Step by step solution

01

Calculate Moles in the Initial Solution

First, to find the number of moles in the initial solution, use the formula for moles: \[ \text{moles} = M \times V \]where \( M \) is the molarity of the solution and \( V \) is the volume in liters. With \( M = 0.110 \text{ M} \) and \( V = 25.0 \text{ mL} = 0.0250 \text{ L} \), the calculation is:\[ \text{moles} = 0.110 \times 0.0250 = 0.00275 \text{ mol} \]
02

Calculate Concentration After First Dilution

The total moles of solute remains the same after the first dilution, but the volume changes. The new concentration \( C_1 \) can be calculated using:\[ C_1 = \frac{\text{moles}}{\text{new volume}} \]where the new volume is \( 100.0 \text{ mL} = 0.1000 \text{ L} \). Plugging in the values from Step 1, the calculation is:\[ C_1 = \frac{0.00275}{0.1000} = 0.0275 \text{ M} \]
03

Determine Moles in Taken Diluted Solution

Next, calculate the amount of moles in the \( 10.0 \text{ mL} \) sample taken from this solution using the concentration \( C_1 \): \[ \text{moles}_{10.0\, \text{mL}} = C_1 \times V = 0.0275 \times 0.0100 = 0.000275 \text{ mol} \]
04

Calculate Final Concentration After Second Dilution

Finally, for the second dilution, use the moles calculated in Step 3 with the new total volume of \( 250 \text{ mL} = 0.250 \text{ L} \) to find the final concentration \( C_f \): \[ C_f = \frac{\text{moles}_{10.0\, \text{mL}}}{0.250} = \frac{0.000275}{0.250} = 0.00110 \text{ M} \]

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Key Concepts

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

Molarity Calculation
Molarity is a fundamental concept in chemistry, helping us to express the concentration of a solute in a solution. The molarity (M) is calculated as the number of moles of solute divided by the volume of solution in liters. This relation is expressed with the formula: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]Understanding and applying this formula is essential for precise solution preparation and accurate chemical reactions.
In the given exercise, you start with a solution concentration where M = 0.110 M, and volume V = 25.0 mL or 0.0250 L. Multiplying these gives the moles: \(0.00275 \text{ moles} \).
When diluting solutions, the number of moles stays constant, but distributions change with volume variations. This implies that knowing initial moles lets you calculate the molarity of any resulting solution, just by adjusting the volume.
This is fundamental to predict how dilutions affect concentration without changing the amount of the solute itself.
Volumetric Flask
A volumetric flask is a piece of laboratory apparatus used for the precise preparation of solutions. Its primary use is to ensure that a solution is diluted to an exact volume. These flasks are uniquely designed with a long narrow neck and a precise volume marking, often used in exercises like the one described to scale solutions accurately. It is crucial for maintaining consistency in experiments and calculations.
When working with a volumetric flask, you add a known quantity of your solute, then fill up with solvent (like distilled water) up to that graduation mark. By doing this, you can ensure that your solution correctly reaches the intended final volume. In the exercise, you start with a smaller volume, including it in a 250 mL volumetric flask, and ultimately fill it up precisely to this mark.
This ensures that every time you replicate this process, the resulting concentration is the same, which is vital for reproducibility in experiments.
Stoichiometry Steps
Stoichiometry is the calculation of reactants and products in chemical reactions. In dilution problems like the given scenario, stoichiometry helps track the quantity changes through each step, guiding problem-solving. The process starts with calculating the initial amount of substance using the molarity formula. For each dilution, you recalculate concentrations, keeping track of moles, which remain constant across dilutions. Using stoichiometry involves a clear understanding of how volumes change affect concentrations, similar to tracking a recipe that changes servings.
The steps in this exercise involve:
  • First, finding initial moles before any dilution begins.
  • Second, calculating the new concentration after the first dilution using the reduced molarity with a larger volume.
  • Then, calculating moles taken from this new solution for second dilution.
  • Finally, recalculating the concentration after filling a larger flask ensures precise stoichiometric steps, optimizing the final concentration for accuracy.
These meticulous calculations are crucial for achieving the correct concentrations for any intended reactions in the lab, assuring correct results.

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Most popular questions from this chapter

What is the mass of solute, in grams, in 125 mL. of a \(1.023 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\) What is the molar concentration of the \(\mathrm{Na}^{+}\) and \(\mathrm{PO}_{4}^{3-}\) ion?

Solution Concentration If \(6.73 \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is dissolved in enough water to make \(250 .\) mL of solution, what is the molar concentration of the sodium carbonate? What are the molar concentrations of the \(\mathrm{Na}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) ions?

A A compound has been isolated that can have either of two possible formulas: (a) \(\mathrm{K}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) or (b) \(\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right] .\) To find which is correct, you dissolve a weighed sample of the compound in acid, forming oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). You then titrate this acid with potassium permanganate, \(\mathrm{KMnO}_{4}\) (the source of the \(\left.\mathrm{MnO}_{4}-\text { ion }\right) .\) The balanced, net ionic equation for the titration is $$ \begin{aligned} 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+& 2 \mathrm{MnO}_{4}-(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow \\ & 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_{2}(\mathrm{g})+14 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$ Titration of \(1.356 \mathrm{g}\) of the compound requires \(34.50 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{KMnO}_{4} .\) Which is the correct formula of the iron-containing compound: (a) or (b)?

Black smokers are found in the depths of the oceans (page 110 ). Thinking that the conditions in these smokers might be conducive to the formation of organic compounds, two chemists in Germany found the following reaction could occur in similar conditions. $$ 2 \mathrm{CH}_{3} \mathrm{SH}+\mathrm{CO} \rightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3}+\mathrm{H}_{2} \mathrm{S} $$ If you begin with \(10.0 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{SH}\) and excess \(\mathrm{CO}\) (a) What is the theoretical yield of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3} ?\) (b) If \(8.65 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3}\) is isolated, what is its percent yield?

A Potassium perchlorate is prepared by the following sequence of reactions: \(\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow\) $$ \mathrm{KCl}(\mathrm{aq})+\mathrm{KClO}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ \(3 \mathrm{KClO}(\mathrm{aq}) \rightarrow 2 \mathrm{KCl}(\mathrm{aq})+\mathrm{KClO}_{3}(\mathrm{aq})\) \(4 \mathrm{KClO}_{3}(\mathrm{aq}) \rightarrow 3 \mathrm{KClO}_{4}(\mathrm{aq})+\mathrm{KCl}(\mathrm{aq})\) What mass of \(\mathrm{Cl}_{2}(\mathrm{g})\) is required to produce \(234 \mathrm{kg}\) of \(\mathrm{KClO}_{4} ?\)

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