/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Calculate the final concentratio... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the final concentrations of the following aqueous solutions after each has been diluted to a final volume of \(25.0 \mathrm{mL}:\) a. \(1.00 \mathrm{mL}\) of \(0.452 M \mathrm{Na}^{+}\) b. \(2.00 \mathrm{mL}\) of \(3.4 \mathrm{m} M \mathrm{LiCl}\) c. \(5.00 \mathrm{mL}\) of \(6.42 \times 10^{-2} \mathrm{m} M \mathrm{Zn}^{2+}\)

Short Answer

Expert verified
a. Na+ ions with an initial concentration of 0.452 M and an initial volume of 1.00 mL. b. LiCl with an initial concentration of 3.40 mM and an initial volume of 2.00 mL. c. Zn2+ ions with an initial concentration of 6.42 x 10^-2 mM and an initial volume of 5.00 mL. Answer: a. The final concentration of Na+ ions after dilution is 0.01808 M. b. The final concentration of LiCl after dilution is 2.72 x 10^-4 M. c. The final concentration of Zn2+ ions after dilution is 1.284 x 10^-2 mM.

Step by step solution

01

a. Final concentration of Na+ ions

In this case, we are given the initial concentration (C1) as 0.452 M and initial volume (V1) as 1.00 mL, and we need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. We can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((0.452 \mathrm{M}) \times (1.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(0.452 \mathrm{M}) \times (1.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 0.01808 \mathrm{M}\) Thus, the final concentration of Na+ ions after dilution is 0.01808 M.
02

b. Final concentration of LiCl

In this case, we are given the initial concentration (C1) as 3.40 mM and initial volume (V1) as 2.00 mL. We need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. Again, we can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((3.4 \times 10^{-3} \mathrm{M}) \times (2.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(3.4 \times 10^{-3} \mathrm{M}) \times (2.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 2.72 \times 10^{-4} \mathrm{M}\) Thus, the final concentration of LiCl after dilution is \(2.72 \times 10^{-4} \mathrm{M}\).
03

c. Final concentration of Zn2+ ions

In this case, we are given the initial concentration (C1) as \(6.42 \times 10^{-2} \mathrm{m} M\) and initial volume (V1) as 5.00 mL. We need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. Again, we can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((6.42 \times 10^{-2} \mathrm{m} M) \times (5.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(6.42 \times 10^{-2} \mathrm{m} M) \times (5.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 1.284 \times 10^{-2} \mathrm{m} M\) Thus, the final concentration of Zn2+ ions after dilution is \(1.284 \times 10^{-2} \mathrm{m} M\).

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Key Concepts

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

Concentration Calculations
Concentration calculations are a foundation in chemistry, used to determine how much of a solute is present within a given volume of a solution. These calculations help us understand the strength or potency of the solution by knowing how concentrated the solute particles are. Imagine you're adding sugar to your coffee; the more sugar you add, the more concentrated your drink becomes. In chemistry, concentration is typically measured in molarity (M), which refers to the number of moles of solute per liter of solution. This helps us gauge how much reactants we might need in a chemical reaction, or what effects a solution might have in practical applications.

To work out concentration, you can use the simple formula: \[ \text{Concentration (M)} = \frac{\text{Amount of Solute (moles)}}{\text{Volume of Solution (liters)}} \] In practice, you will often need to convert volumes and amounts into moles and liters. This steady approach gives insights into how strong or weak a solution is, preparing you to observe and predict chemical behaviors accurately.
Molarity and Solution Preparation
Molarity is an integral component of solution preparation in chemistry. It's defined as the moles of solute divided by the liters of solution, and it helps chemists to accurately quantify the components of a solution. Preparing a solution with a specific molarity is essential when performing experiments because it ensures consistency and repeatability.

Here's the basic approach to preparing a solution:
  • Determine the desired molarity of the solution.
  • Calculate the number of moles of solute needed using the formula: \( \text{Moles of solute} = \text{Molarity} \times \text{Volume in liters} \).
  • Dissolve the calculated amount of solute in a solvent, usually water, and adjust the total volume to the required amount.
This meticulous method allows for the preparation of solutions with precise concentrations, ensuring that each experimental setup remains consistent.
Dilution Formula
The dilution formula is a handy tool when you need to make a less concentrated solution from a more concentrated one. This is particularly useful when you have limited resources or need specific concentrations for various experiments.

The formula used for dilution is: \[ C_1 \times V_1 = C_2 \times V_2 \] Where:
  • \( C_1 \) is the initial concentration of the solution.
  • \( V_1 \) is the initial volume of the solution.
  • \( C_2 \) is the final concentration after dilution.
  • \( V_2 \) is the final total volume after dilution.
To use this formula, balance the product of the initial concentration and volume with the new, diluted concentration and final volume. This equation assumes a conservative system where the quantity of solute remains constant, but the solution volume changes. It's a straightforward approach that allows chemists to determine the new concentration quickly.

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Most popular questions from this chapter

Many nonmetal oxides react with water to form acidic solutions. Give the formulas of the acids produced in the following reactions: a. \(P_{4} O_{10}+6 H_{2} O \rightarrow ?\) b. \(\mathrm{SeO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow ?\) c. \(\mathrm{B}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow ?\)

HF is prepared by reacting \(\mathrm{CaF}_{2}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}:\) $$ \mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \rightarrow 2 \mathrm{HF}(g)+\mathrm{CaSO}_{4}(s) $$ HF can be electrolyzed, in turn, when dissolved in molten KF to produce fluorine gas: $$ 2 \mathrm{HF}(\ell) \rightarrow \mathrm{F}_{2}(g)+\mathrm{H}_{2}(g) $$ Fluorine is extremely reactive, so it is typically sold as a \(5 \%\) mixture by volume in an inert gas such as helium. How much \(\mathrm{CaF}_{2}\) is required to produce \(500.0 \mathrm{L}\) of \(5 \% \mathrm{F}_{2}\) in helium? Assume the density of \(\mathrm{F}_{2}\) gas is \(1.70 \mathrm{g} / \mathrm{L}\)

Some people who prefer natural foods make their own apple cider vinegar. They start with freshly squeezed apple juice that contains about \(6 \%\) natural sugars. These sugars, which all have nearly the same empirical formula, \(\mathrm{CH}_{2} \mathrm{O},\) are fermented with yeast in a chemical reaction that produces equal numbers of moles of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) and carbon dioxide. The product of fermentation, called hard cider, undergoes an acid fermentation step in which ethanol and dissolved oxygen gas react together to form acetic acid (CH \(_{3} \mathrm{COOH}\) ) and water. This acetic acid is the principal solute in vinegar. a. Write a balanced chemical equation for the fermentation of natural sugars to ethanol and carbon dioxide. You may use in the equation the empirical formula given in the preceding paragraph. b. Write a balanced chemical equation for the acid fermentation of ethanol to acetic acid. c. What are the oxidation states of carbon in the reactants and products of the two fermentation reactions? d. If a sample of apple juice contains \(1.00 \times 10^{2} \mathrm{g}\) of natural sugar, what is the maximum quantity of acetic acid that could be produced by the two fermentation reactions?

Calcium nitrate is soluble, but calcium phosphate is not. Explain this difference in solubility

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, and write the net ionic equations: a. \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) b. \(\mathrm{NiCl}_{2}(a q)+\mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \rightarrow\) c. \(\operatorname{Fe} C l_{2}(a q)+N a_{2} S(a q) \rightarrow\) d. \(\operatorname{MgSO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow\)

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