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Chapter 4: Problem 80

You want to analyze a silver nitrate solution. What mass of \(\mathrm{NaCl}\) is needed to precipitate \(\mathrm{Ag}^{+}\) ions from \(45.0 \mathrm{~mL}\) of \(0.2500 \mathrm{MAgNO}_{3}\) solution?

Short Answer

Expert verified
To precipitate all the Ag⁺ ions from the 45.0 mL of 0.2500 M AgNO₃ solution, approximately 0.6574 grams of NaCl are needed.

Step by step solution

01

Write the chemical equation

The chemical reaction between silver nitrate (AgNO₃) and sodium chloride (NaCl) produces a precipitate of silver chloride (AgCl) and sodium nitrate (NaNO₃). The balanced chemical equation is: AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
02

Determine the moles of Ag⁺ ions in the solution

Using the given volume and concentration of the silver nitrate (AgNO₃) solution, we can calculate the moles of Ag⁺ ions present: Moles of Ag⁺ = Volume * Concentration Moles of Ag⁺ = 45.0 mL * 0.2500 M Keep in mind that we need to convert the volume from mL to L before multiplying with the concentration: Moles of Ag⁺ = 0.045 L * 0.2500 M Moles of Ag⁺ = 0.01125 mol
03

Determine the moles of NaCl needed using stoichiometry

Using the balanced chemical equation, we can determine the moles of NaCl needed to precipitate all the Ag⁺ ions: 1 mol AgNO₃ : 1 mol NaCl Given that there are 0.01125 mol of Ag⁺ ions, the required moles of NaCl will be equal: Moles of NaCl = 0.01125 mol
04

Calculate the mass of NaCl required

Using the molar mass of NaCl, we can calculate the mass of NaCl needed to precipitate all the Ag⁺ ions: Molar mass of NaCl = 58.44 g/mol Mass of NaCl = Moles of NaCl * Molar mass of NaCl Mass of NaCl = 0.01125 mol * 58.44 g/mol Mass of NaCl = 0.6574 g Therefore, approximately 0.6574 grams of NaCl are needed to precipitate all the Ag⁺ ions from the 45.0 mL of 0.2500 M AgNO₃ solution.

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Most popular questions from this chapter

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

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