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Chapter 4: Problem 81

A \(3.50 \mathrm{~g}\) sample of \(\mathrm{KCl}\) is dissolved in \(10.0 \mathrm{~mL}\) of water. The resulting solution is then added to \(60.0 \mathrm{~mL}\) of a \(0.500 \mathrm{M}\) \(\mathrm{CaCl}_{2}(a q)\) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution.

Short Answer

Expert verified
[K鈦篯 = 0.670 M, [Ca虏鈦篯 = 0.429 M, [Cl鈦籡 = 1.53 M.

Step by step solution

01

Calculate moles of KCl

First, we need to determine the moles of KCl present in the 3.50 gram sample. The molar mass of KCl is approximately 74.55 g/mol.\[ \text{Moles of } \mathrm{KCl} = \frac{3.50\, \text{g}}{74.55\, \text{g/mol}} \approx 0.0469\, \text{mol} \]
02

Determine volume of final solution

Determine the total volume of the solution after mixing. The original 10.0 mL from dissolving KCl and the 60.0 mL of CaClelement in solution:\[ V_{\text{final}} = 10.0\, \text{mL} + 60.0\, \text{mL} = 70.0\, \text{mL} \]
03

Calculate concentration of K+ ion

Each mole of KCl yields one K鈦� ion when dissolved. Since the total final volume of the solution is 70.0 mL:\[ [\mathrm{K}^+] = \frac{0.0469\, \text{mol}}{0.070\, \text{L}} = 0.670\, \text{M} \]
04

Calculate concentration of Cl- ion from KCl

Similarly, each mole of KCl also yields one Cl鈦� ion:\[ [\mathrm{Cl}^-]_{\mathrm{KCl}} = \frac{0.0469\, \text{mol}}{0.070\, \text{L}} = 0.670\, \text{M} \]
05

Calculate moles of CaCl2

Calculate the moles of CaCl鈧� in the original solution. Given concentration is 0.500 M and the initial volume is 60.0 mL (0.060 L):\[ \text{Moles of } \mathrm{CaCl}_2 = 0.500\, \text{M} \times 0.060\, \text{L} = 0.0300\, \text{mol} \]
06

Calculate concentration of Ca2+ ion

Since each mole of CaCl鈧� provides one Ca虏鈦� ion, the concentration is:\[ [\mathrm{Ca}^{2+}] = \frac{0.0300\, \text{mol}}{0.070\, \text{L}} = 0.429\, \text{M} \]
07

Determine total Cl- ion concentration

The Cl鈦� ions come from both KCl and CaCl鈧�. Since each mole of CaCl鈧� gives two Cl鈦� ions, the moles of Cl鈦� from CaCl鈧� is:\[ \text{Moles of } \mathrm{Cl}^- = 2 \times 0.0300\, \text{mol} = 0.0600\, \text{mol} \]Adding the moles from KCl:\[ \text{Total moles of } \mathrm{Cl}^- = 0.0469\, \text{mol} + 0.0600\, \text{mol} = 0.1069\, \text{mol} \]
08

Calculate final Cl- concentration

The concentration of Cl鈦� is then:\[ [\mathrm{Cl}^-] = \frac{0.1069\, \text{mol}}{0.070\, \text{L}} = 1.53\, \text{M} \]

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Key Concepts

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

Ion Concentration Calculation
In chemistry, calculating ion concentration is an essential skill when analyzing solutions. This process involves determining the number of moles of ions in a solution and then dividing by the total volume of the solution, turning it into molarity. When a substance like KCl is dissolved in water, K鈦� and Cl鈦� ions are released. It's vital to separately compute the moles of each ion based on the compound's dissociation.

For example, if you dissolve a sample of KCl, calculate the moles using its molar mass. Then, use this mole count to find out each ion's concentration:
  • For KCl, each mole of the compound results in one mole of K鈦� and one mole of Cl鈦�.
  • The concentration of each ion is simply the moles of the ion divided by the total volume of the solution in liters.
Understanding these granular steps helps you to analyze and predict the chemical behavior of solutions, especially when multiple substances are mixed together.
Molarity
Molarity (M) is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the liters of solution. Molarity is central in chemistry for quantifying the concentration without ambiguity about the quantity of solvent.

To calculate molarity:
  • First, determine the moles of the solute, typically from its mass and molar mass.
  • Second, know the total volume of the solution in liters.
  • Finally, use the formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).
For example, a 0.500 M CaCl鈧� solution means there are 0.5 moles of CaCl鈧� per liter of solution. Molarity is crucial when preparing solutions in laboratories, as it directly impacts reaction stoichiometry and overall outcomes.
Chemical Mixing
When mixing chemicals in a solution, the volumes are often assumed to be additive, meaning the total volume is the sum of the individual volumes. This assumption simplifies calculations and is usually valid for dilute solutions.

In our exercise, KCl is dissolved in water, and this solution is added to another, resulting in a total volume of 70 mL. While this might seem straightforward, concentration calculations become complex due to differing initial concentrations, different ions introduced, and their subsequent dissociation.
  • Always calculate the initial concentrations and volumes before mixing.
  • Track each ion individually to correctly calculate the final concentrations.
Understanding chemical mixing ensures that the properties of the final solution are known and controllable, which is crucial in both experimental design and industrial processes.

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Most popular questions from this chapter

Classify each of the following reactions as a combination reaction, decomposition reaction, displacement reaction, or combustion reaction. a. When they are heated, ammonium dichromate crystals, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), decompose to give nitrogen, water vapor, and solid chromium(III) oxide, \(\mathrm{Cr}_{2} \mathrm{O}_{3}\). b. When aqueous ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), is heated, it gives nitrogen and water vapor. C. When gaseous ammonia, \(\mathrm{NH}_{3}\), reacts with hydrogen chloride gas, HCl, fine crystals of ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), are formed. d. Aluminum added to an aqueous solution of sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), forms a solution of aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) Hydrogen gas is released.

In the following reactions, label the oxidizing agent and the reducing agent. a. \(\mathrm{ZnO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Zn}(g)+\mathrm{CO}(g)\) b. \(8 \mathrm{Fe}(s)+\mathrm{S}_{8}(s) \longrightarrow 8 \mathrm{FeS}(s)\)

A sample of \(0.0512\) mol of iron(III) chloride, \(\mathrm{FeCl}_{3}\), was dissolved in water to give \(25.0 \mathrm{~mL}\) of solution. What is the molarity of the solution?

A \(3.33-\mathrm{g}\) sample of iron ore is transformed to a solution of iron(II) sulfate, \(\mathrm{FeSO}_{4}\), and this solution is titrated with \(0.150 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (potassium dichromate). If it requires \(43.7 \mathrm{~mL}\) of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is $$ \begin{aligned} &6 \mathrm{FeSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ &3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+7 \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{aligned} $$

In the following reactions, label the oxidizing agent and the reducing agent. a. \(2 \mathrm{Al}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{AlF}_{3}(s)\) b. \(\mathrm{Hg}^{2+}(a q)+\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow\) $$ \mathrm{Hg}(s)+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) $$
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