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Chapter 4: Problem 67

A \(35.2-\mathrm{mL}, 1.66 \mathrm{M} \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

Short Answer

Expert verified
The final concentration of the solution is approximately 1.413 M.

Step by step solution

01

Calculate moles of KMnO4 in the first solution

Use the formula for molarity, which is \( M = \frac{n}{V} \). Rearrange it to find the number of moles \( n \) in the first solution: \[ n_1 = M_1 \times V_1 \]Where:- \( M_1 = 1.66 \; \text{M} \)- \( V_1 = 35.2 \; \text{mL} = 0.0352 \; \text{L} \)Substitute the values:\[ n_1 = 1.66 \times 0.0352 = 0.058432 \; \text{moles} \]
02

Calculate moles of KMnO4 in the second solution

Similarly, calculate the number of moles in the second solution using the formula:\[ n_2 = M_2 \times V_2 \]Where:- \( M_2 = 0.892 \; \text{M} \)- \( V_2 = 16.7 \; \text{mL} = 0.0167 \; \text{L} \)Substitute the values:\[ n_2 = 0.892 \times 0.0167 = 0.0148964 \; \text{moles} \]
03

Calculate total moles of KMnO4 in the mixture

To find the total moles in the final solution, add the moles from both solutions:\[ n_{\text{total}} = n_1 + n_2 \]Substituting the values from the previous steps:\[ n_{\text{total}} = 0.058432 + 0.0148964 = 0.0733284 \; \text{moles} \]
04

Calculate total volume of the mixture

Add the volumes of both solutions to find the total volume of the mixture:\[ V_{\text{total}} = V_1 + V_2 \]Substitute the volumes:\[ V_{\text{total}} = 0.0352 + 0.0167 = 0.0519 \; \text{L} \]
05

Calculate concentration of the final solution

Now use the formula for molarity with the total moles and total volume:\[ M_{\text{final}} = \frac{n_{\text{total}}}{V_{\text{total}}} \]Substitute the calculated values:\[ M_{\text{final}} = \frac{0.0733284}{0.0519} \approx 1.413 \; \text{M} \]

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Key Concepts

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

Understanding Molarity Formula
In chemistry, molarity is a key measure to express the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. This formula helps chemists determine how much of a substance is present in a given volume:\[M = \frac{n}{V}\]Where:
  • M is the molarity,
  • n is the number of moles of the solute,
  • V is the volume of the solution in liters.
The molarity formula is straightforward, yet its accuracy depends on correctly measuring volumes and calculating moles. By using this formula, one can solve for the unknown if the other two quantities are known. For example, in the original problem, we calculated the number of moles of KMnO4 using the solutions' molarity and volumes. Understanding how to manipulate this formula is crucial for solving problems involving solution concentrations.
Chemical Mixtures and Their Properties
When two or more solutions are combined, they create a chemical mixture. In these mixtures, the substances are physically intermingled, but they do not chemically react with each other. Mixing solutions involves combining their solute amounts and volumes to find the resultant concentration.

In the given exercise, we mixed two different KMnO4 solutions. Each solution initially had its concentration as given by its molarity. By adding them together, we had to calculate the total number of moles present and the total volume of the mixture. These steps are crucial:
  • Calculate the moles for each solution separately using their molarity and volume.
  • Add the number of moles together for the total moles in the mixture.
  • Combine the volumes of the separate solutions to determine the total volume.
By understanding these steps, we can determine the concentration of the new mixture using the molarity formula once again. Keeping these properties in mind helps in understanding how different solutions can be combined and how their final properties can be determined.
The Role of Stoichiometry in Mixtures
Stoichiometry comes into play whenever dealing with the quantitative relationships in chemical mixtures. It ensures the exact amount of reactants and products is considered in chemical reactions. However, in non-reactive processes like mixing solutions, stoichiometry helps in calculating accurate proportions of substances.

In the exercise, while stoichiometry is not about reacting substances, it helps calculate the precise amounts of substances in the mixture. It involves steps such as:
  • Determining the moles of each solution using their molarity and volumes.
  • Adding these moles to find the total number of moles in the mixture, which is crucial for finding the mixture's overall concentration.
Stoichiometry is essentially a bookkeeping method for atoms and molecules in chemistry, allowing chemists to predict outcomes of reactions and process efficiencies. Though our specific case didn't involve a chemical reaction, stoichiometry aided in meticulously calculating the end result, ensuring our final solution concentration was accurate. Understanding stoichiometry in various contexts broadens its application beyond mere chemical reactions.

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Most popular questions from this chapter

Water is an extremely weak electrolyte and therefore cannot conduct electricity. Why are we often cautioned not to operate electrical appliances when our hands are wet?

Which of these aqueous solutions would you expect to be the best conductor of electricity at \(25^{\circ} \mathrm{C} ?\) Explain your answer. (a) \(0.20 M \mathrm{NaCl}\) (b) \(0.60 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (c) \(0.25 M \mathrm{HCl}\) (d) \(0.20 M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Distilled water must be used in the gravimetric analysis of chlorides. Why?

The concentration of \(\mathrm{Cu}^{2+}\) ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide \(\left(\mathrm{Na}_{2} \mathrm{~S}\right)\) solution to \(0.800 \mathrm{~L}\) of the water. The molecular equation is $$ \mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CuS}(s) $$ Write the net ionic equation and calculate the molar concentration of \(\mathrm{Cu}^{2+}\) in the water sample if \(0.0177 \mathrm{~g}\) of solid CuS is formed.

Magnesium is a valuable, lightweight metal. It is used as a structural metal and in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth's crust, it is cheaper to "mine" the metal from seawater. Magnesium forms the second most abundant cation in the sea (after sodium); there are about \(1.3 \mathrm{~g}\) of magnesium in \(1 \mathrm{~kg}\) of seawater. The method of obtaining magnesium from seawater employs all three types of reactions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is heated at high temperatures to produce quicklime, or calcium oxide \((\mathrm{CaO})\) : $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ When calcium oxide is treated with seawater, it forms calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right],\) which is slightly soluble and ionizes to give \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions: $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ The surplus hydroxide ions cause the much less soluble magnesium hydroxide to precipitate: $$ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) $$ The solid magnesium hydroxide is filtered and reacted with hydrochloric acid to form magnesium chloride \(\left(\mathrm{MgCl}_{2}\right):\) $$ \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \underset{\mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)}{\longrightarrow} $$ After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are $$ \begin{aligned} \mathrm{Mg}^{2+}+2 e^{-} \longrightarrow & \mathrm{Mg} \\ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-} \end{aligned} $$ The overall reaction is $$ \mathrm{MgCl}_{2}(l) \longrightarrow \mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) $$ This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloric acid and recycled through the process. (a) Identify the precipitation, acid-base, and redox processes. (b) Instead of calcium oxide, why don't we simply add sodium hydroxide to precipitate magnesium hydroxide? (c) Sometimes a mineral called dolomite (a combination of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) ) is substituted for limestone \(\left(\mathrm{CaCO}_{3}\right)\) to bring about the precipitation of magnesium hydroxide. What is the advantage of using dolomite? (d) What are the advantages of mining magnesium from the ocean rather than from Earth's crust?
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