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What volume of \(0.248 \mathrm{M} \mathrm{CaCl}_{2}\) must be added to \(335 \mathrm{mL}\) of \(0.186 \mathrm{M} \mathrm{KCl}\) to produce a solution with a concentration of \(0.250 \mathrm{M} \mathrm{Cl}^{-2}\) Assume that the solution volumes are additive.

Short Answer

Expert verified
You need to add 94.7 mL of 0.248M CaCl2 solution to the 335 mL of 0.186M KCl solution to produce a solution with a Cl- concentration of 0.250M.

Step by step solution

01

Calculate moles of chloride in KCl solution

Moles of chloride in KCl is given by the product of volume and molarity, since KCl has only one atom of chloride, so we have \((0.186 Moles/L)*(0.335L) = 0.06231 Moles\)
02

Calculate total moles of chloride required

The total moles of Cl- required to achieve 0.250M in the entire solution. We do not know the final volume yet, but it will be 335 mL of the KCl solution + X mL of the CaCl2 solution. As both these volumes are in mL, they can be added directly, This gives us \(Final Moles of Cl- = 0.250 M * (335+X) mL * 1L/1000mL\)
03

Determine moles of chloride in CaCl2

The total moles of Cl- required (from step 2) will be equal to the moles contributed by the KCl solution and the CaCl2 solution. And each molecule of CaCl2 contributes 2 Cl- ions. This gives us \(0.250 M * (335+X) mL * 1L/1000mL = 0.06231 Moles(from KCl) + 2 * 0.248M * X mL * 1L/1000mL(from CaCl2)\)
04

Solve for X (volume of CaCl2 solution)

This is a simple algebra problem now. The only unknown in the equation is X - the volume in mL of the CaCl2 solution. Solving the equation for X gives \(X = 94.7 mL\)

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Key Concepts

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

Molarity
Molarity is a way to express the concentration of a solution, showing how many moles of a solute are present per liter of solution. Understanding molarity helps you calculate and prepare solutions of specific concentrations with precision. Here's a friendly breakdown:
  • It's expressed in moles per liter (mol/L), represented by the symbol "M".
  • A 1 M solution has 1 mole of a solute per 1 liter of solution. For example, a 0.248 M CaCl_2 solution means there are 0.248 moles of calcium chloride in each liter.
Molarity is vital for stoichiometry in chemical reactions and conducting experiments. When combining solutions, as in our exercise, knowing the molarity allows us to predict and control the exact outcome as desired.
Additive Volumes
Additive volumes refer to the assumption that when two solutions are mixed, the total volume is simply the sum of the individual volumes. This isn't always precisely accurate in real-world scenarios because of interactions between molecules, but it's a common simplification in chemistry calculations.
  • For our exercise, mixing 335 mL of KCl solution and a certain volume of CaCl_2, the total volume is assumed to be the sum of both.
  • This simplification helps in concentration calculations by providing an easy way to determine the final volume.
While additive volumes simplify calculations, remember that in highly precise fields, real measurements might show slight deviations.
Concentration Calculations
Concentration calculations involve determining the amount of solute present in a given volume of solution. This process often requires combining several mathematical steps to find a desired concentration, like achieving a specific chloride ion concentration in the exercise.
  • First, calculate the moles of solute already present, using molarity and existing solution volumes.
  • Determine the desired concentration outcome after adding more solution.
  • Solve for unknowns, such as additional volumes required, ensuring the concentration target is met.
Understanding these calculations ensures that you mix solutions correctly for desired chemical compositions in educational and practical applications, making it a fundamental skill in chemistry.

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Most popular questions from this chapter

In your own words, define or explain the terms or symbols \((\mathrm{a}) \rightleftharpoons(\mathrm{b})[] ;(\mathrm{c})\) spectator ion; (d) weak acid.

An unknown white solid consists of two compounds, each containing a different cation. As suggested in the illustration, the unknown is partially soluble in water. The solution is treated with \(\mathrm{NaOH}(\mathrm{aq})\) and yields a white precipitate. The part of the original solid that is insoluble in water dissolves in \(\mathrm{HCl}(\mathrm{aq})\) with the evolution of a gas. The resulting solution is then treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and yields a white precipitate. (a) Is it possible that any of the cations \(M g^{2+}, C u^{2+}\) \(\mathrm{Ba}^{2+}, \mathrm{Na}^{+},\) or \(\mathrm{NH}_{4}^{+}\) were present in the original unknown? Explain your reasoning. (b) What compounds could be in the unknown mixture (that is, what anions might be present)?

In the half-reaction in which \(\mathrm{NpO}_{2}^{+}\) is converted to \(\mathrm{Np}^{4+},\) the number of electrons appearing in the half-equation is (a) \(1 ;(b) 2 ;(c) 3 ;\) (d) 4.

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

The active ingredients in a particular antacid tablet are aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2} . \quad \mathrm{A} 5.00 \times 10^{2} \mathrm{mg}\) sample of the active ingredients was dissolved in \(50.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl} .\) The resulting solution, which was still acidic, required \(16.5 \mathrm{mL}\) of \(0.377 \mathrm{M} \mathrm{NaOH}\) for neutralization. What are the mass percentages of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the sample?

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