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A \(46.2-\mathrm{mL}, 0.568 M\) calcium nitrate \(\left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) solution is mixed with \(80.5 \mathrm{~mL}\) of \(1.396 \mathrm{M}\) calcium nitrate solution. Calculate the concentration of the final solution.

Short Answer

Expert verified
The concentration of the final solution is \(1.094 M\).

Step by step solution

01

Calculate the total moles of calcium nitrate in each solution

Use the formula \(M = n/V\) to find the moles of calcium nitrate in each solution, where \(M\) is molarity, \(n\) is the number of moles, and \(V\) is the volume. Therefore, for the first solution, the moles are \(n_1 = M_1V_1 = 0.568 M * 46.2 mL = 26.22 mmol\), and for the second solution, the moles are \(n_2 = M_2V_2 = 1.396 M * 80.5 mL = 112.38 mmol\). Note that \(M = mol/L\) and that the volumes are in mL, so the number of moles is in mmol.
02

Determine the total moles of calcium nitrate

Combine the moles of calcium nitrate in both solutions to get the total moles. So, \(n_t = n_1 + n_2 = 26.22 mmol + 112.38 mmol = 138.6 mmol\).
03

Find the total volume of the mixed solutions

Add the volumes of the initial solutions to get the total volume. Hence, \(V_t = V_1 + V_2 = 46.2 mL + 80.5mL = 126.7 mL\).
04

Calculate the final concentration

Use the formula for molarity, \(M = n/V\), to get the final concentration. Substitute the total moles and the total volume to get the final concentration. Therefore, \(M_f = n_t/V_t = 138.6 mmol / 126.7 mL = 1.094 M\).

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Key Concepts

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

Molarity Formula
Understanding the molarity formula is crucial for calculating the concentration of solutions. Molarity, denoted by the symbol 'M,' is a measure of concentration that tells us how many moles of a solute are present in a liter of solution. The formula to calculate molarity is given as:
\[ M = \frac{n}{V} \]
where 'M' is the molarity, 'n' represents the number of moles of the solute, and 'V' is the volume of the solution in liters. When volume is expressed in milliliters (mL), we can still use this formula by converting mL to L, where 1 L = 1000 mL.
To calculate the moles, you need to know the quantity of the substance in grams and its molar mass. After finding the number of moles, you simply divide by the volume of the solution to find the molarity. This formula is pivotal for mixing solutions to achieve desired concentrations, as seen in our exercise problem where two calcium nitrate solutions with different molarities are mixed.
Concentration of Solutions
The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent. There are various ways to express concentrations, but molarity is one of the most common methods used in chemistry due to its straightforward relationship with the volume of the solution.
Molarity is especially helpful because it allows chemists to easily calculate the amount of a substance in a solution, necessary for reactions and dilutions. When two solutions with known molarities are mixed, the concentration of the final solution can be found by adding the moles of solute from both original solutions and dividing by the total volume, as demonstrated in the provided solution.
This concept is not just limited to academic exercises but also essential in real-life applications, such as formulating medicines, calibrating pH solutions, or creating standard solutions for laboratory experiments.
Moles and Volume Relationship
The relationship between moles and volume is a fundamental aspect of chemical solutions. The number of moles (\(n\)) is a measure of the quantity of particles (atoms, molecules, ions, etc.) and can be derived from the mass of the substance and its molar mass. Volume (\(V\)), particularly in the context of solutions, is typically measured in liters or milliliters.
When calculating the final concentration of a mixed solution, understanding this relationship is essential. It's all about balancing the number of moles with the resultant volume. For the individual components before mixing, the molarity (moles per volume) gives us a clear picture of their concentrations. After mixing, it's the total number of moles divided by the total volume that yields the final concentration.
Using the problem at hand, the exercises' advised approach in combining the moles from both solutions and then the total volume highlights how to manipulate these quantities to arrive at a consolidated concentration for a new, combined solution.

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Most popular questions from this chapter

Identify the following as a weak or strong acid or base: (a) \(\mathrm{NH}_{3},\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4},\) (c) \(\mathrm{LiOH},\) d) \(\mathrm{HCOOH}\) (formic acid), (e) \(\mathrm{H}_{2} \mathrm{SO}_{4},\) (f) \(\mathrm{HF},(\mathrm{g}) \mathrm{Ba}(\mathrm{OH})_{2}\)

A 5.012 -g sample of an iron chloride hydrate was dried in an oven. The mass of the anhydrous compound was \(3.195 \mathrm{~g}\). The compound was dissolved in water and reacted with an excess of \(\mathrm{AgNO}_{3}\). The precipitae of \(\mathrm{AgCl}\) formed weighed \(7.225 \mathrm{~g}\). What is the formula of the original compound?

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 \(\left(\mathrm{MgCl}_{2}\right)\) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ 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 halfreactions 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?

Calculate the volume in \(\mathrm{mL}\) of a \(1.420 \mathrm{M} \mathrm{NaOH}\) solution required to titrate the following solutions: (a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution (b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution (c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution

If \(30.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) is added to \(15.0 \mathrm{~mL}\) of \(0.100 \mathrm{MAgNO}_{3}\), what is the mass in grams of \(\mathrm{AgCl}\) precipitate?

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