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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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Most popular questions from this chapter

What is the net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

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