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Chapter 5: Problem 17

Assuming the volumes are additive, what is the \(\left[\mathrm{Cl}^{-}\right]\) in a solution obtained by mixing \(225 \mathrm{mL}\) of \(0.625 \mathrm{M}\) \(\mathrm{KCl}\) and \(615 \mathrm{mL}\) of \(0.385 \mathrm{M} \mathrm{MgCl}_{2} ?\)

Short Answer

Expert verified
The concentration of Chloride ions [Cl-] in the resulting solution would be approximately 0.7314 M.

Step by step solution

01

Understand The Chemical Compounds

The compound KCl dissociates in solution to form K+ and Cl- ions. On the other hand, MgCl2 gives us Mg2+ and two Cl- ions in solution. This means that a mole of KCl will have the same number of moles of Cl- ions while a mole of MgCl2 will have twice the number of moles of Cl- ions.
02

Calculate The Moles

For KCl, the number of moles can be calculated by multiplying the volume of the KCl solution (in liters) by its molar concentration. Thus, moles of Cl- from KCl = 225 mL * (1L/1000 mL) * 0.625 M = 0.140625 moles. We can obtain the moles of chloride ions from MgCl2 in the same way: moles of Cl- from MgCl2 = 615 mL * (1 L/1000 mL) * 0.385 M * 2 = 0.47355 moles. The multiplication by 2 accounts for the fact that each molecular unit of MgCl2 gives 2 ions of Cl-.
03

Add The Moles

Now we need to add the moles of Cl- ions from both solutions to get the total moles. Total moles of Cl- = moles of Cl- from KCl + moles of Cl- from MgCl2 = 0.140625 moles + 0.47355 moles = 0.614175 moles.
04

Calculate The Concentration

Finally, the molar concentration (M) of Cl- ions can be obtained by dividing the total moles by the total volume. The total volume = 225 mL + 615 mL = 840 mL = 0.84 L. Therefore, molar concentration (M) = total moles / total volume in Liters = 0.614175 moles / 0.84 L = 0.7314 M.

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Key Concepts

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

Chloride Ion Concentration
Understanding chloride ion concentration in a solution involves recognizing how different compounds contribute their ions when dissolved. In this exercise, we mix solutions of KCl and MgCl鈧�. Both compounds dissolve and dissociate in water to release chloride ions (Cl鈦�).
  • KCl: Each molecule of potassium chloride (KCl) dissociates into one K鈦� ion and one Cl鈦� ion.
  • 惭驳颁濒鈧�: Magnesium chloride (MgCl鈧�) behaves differently. It splits into one Mg虏鈦� ion and two Cl鈦� ions. Thus, MgCl鈧� contributes twice the number of Cl鈦� ions per molecule compared to KCl.
Careful consideration of the dissociation process is key as it influences how many chloride ions are present in the final solution. Knowing the behavior of these molecules upon dissolving helps determine the total chloride ion concentration.
Molarity Calculation
Calculating molarity, which measures concentration, is essential when mixing solutions. Molarity is defined as the number of moles of solute per liter of solution. To find it, we need:
  • Total moles of solute (chloride ions in this case)
  • Volume of the solution in liters
For example, with KCl's molarity known and its volume converted to liters, we calculate the moles of Cl鈦� ions. The same process applies to MgCl鈧�, multiplying by 2 for its two chloride ions per formula unit. Finally, total moles of chloride ions are found by adding the moles from both solutions. Dividing this total by the solution's overall volume gives molarity. This calculation underscores how solute and solvent quantities relate to the concentration of the chloride ions.
Chemical Dissociation
Chemical dissociation refers to the process where ionic compounds separate into individual ions in solution. This understanding is vital for predicting the behavior of dissolved substances.
  • When KCl dissociates, it directly forms one Cl鈦� ion per molecule.
  • However, MgCl鈧�'s dissociation results in two Cl鈦� ions per molecule, doubling its contribution of chloride ions relative to KCl.
Knowing how compounds dissociate allows for accurate calculation of ion concentrations in solutions. This principle not only helps in chemistry exercises but also in real-world applications like determining pollutant concentrations in water supplies or medication dosages. By grasping the basics of dissociation, we understand the fundamental chemical behaviors necessary for calculating subsequent concentrations such as molarity.

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Most popular questions from this chapter

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

A \(0.4324 \mathrm{g}\) sample of a potassium hydroxide-lithium hydroxide mixture requires \(28.28 \mathrm{mL}\) of \(0.3520 \mathrm{M} \mathrm{HCl}\) for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

A sample of battery acid is to be analyzed for its sulfuric acid content. A \(1.00 \mathrm{mL}\) sample weighs \(1.239 \mathrm{g}\). This \(1.00 \mathrm{mL}\) sample is diluted to \(250.0 \mathrm{mL}\), and \(10.00 \mathrm{mL}\) of this diluted acid requires \(32.44 \mathrm{mL}\) of \(0.00498 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the battery acid? (Assume that complete ionization and neutralization of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) occurs.)

Which aqueous solution has the greatest \(\left[\mathrm{H}^{+}\right]:\) (a) \(0.011 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (b) \(0.010 \mathrm{M} \mathrm{HCl} ;\) (c) \(0.010 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(1.00 \mathrm{M} \mathrm{NH}_{3} ?\) Explain your choice.

Explain the important distinctions between (a) a strong electrolyte and strong acid; (b) an oxidizing agent and reducing agent; (c) precipitation reactions and neutralization reactions; (d) half-reaction and overall reaction.
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