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Chapter 5: Problem 75

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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Most popular questions from this chapter

Describe the synthesis of each of the following ionic compounds, starting from solutions of sodium and nitrate salts. Then write the net ionic equation for each synthesis. (a) \(\mathrm{Zn}_{3}\left(\mathrm{PO}_{4}\right)_{2};\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2};\) (c) \(\mathrm{NiCO}_{3}.\)

Determine which of the following react(s) with HCl(ag) to produce a gas, and write a net ionic equation(s) for the reaction(s). (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4} ;\) (b) \(\mathrm{KHSO}_{3}\); (c) \(\mathrm{Zn}(\mathrm{OH})_{2};\) (d) \(\mathrm{CaCl}_{2}\).

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

An \(\mathrm{NaOH}(\mathrm{aq})\) solution cannot be made up to an exact concentration simply by weighing out the required mass of NaOH, because the NaOH is not pure. Also, water vapor condenses on the solid as it is being weighed. The solution must be standardized by titration. For this purpose, a 25.00 \(\mathrm{mL}\) sample of an NaOH(aq) solution requires 28.34 \(\mathrm{mL}\) of 0.1085 \(\mathrm{M}\) HCl. What is the molarity of the NaOH(aq)? \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

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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