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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鈦� ions from the 45.0 mL of 0.2500 M AgNO鈧� solution, approximately 0.6574 grams of NaCl are needed.

Step by step solution

01

Write the chemical equation

The chemical reaction between silver nitrate (础驳狈翱鈧�) and sodium chloride (NaCl) produces a precipitate of silver chloride (AgCl) and sodium nitrate (NaNO鈧�). The balanced chemical equation is: AgNO鈧� (aq) + NaCl (aq) 鈫� AgCl (s) + NaNO鈧� (aq)
02

Determine the moles of Ag鈦� ions in the solution

Using the given volume and concentration of the silver nitrate (础驳狈翱鈧�) solution, we can calculate the moles of Ag鈦� ions present: Moles of Ag鈦� = Volume * Concentration Moles of Ag鈦� = 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鈦� = 0.045 L * 0.2500 M Moles of Ag鈦� = 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鈦� ions: 1 mol AgNO鈧� : 1 mol NaCl Given that there are 0.01125 mol of Ag鈦� 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鈦� 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鈦� ions from the 45.0 mL of 0.2500 M AgNO鈧� solution.

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Key Concepts

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

Chemical Equations
Chemical equations are representations of chemical reactions where reactants transform into products. In this exercise, the chemical reaction involves silver nitrate
  • (础驳狈翱鈧�)
  • and sodium chloride (NaCl).
Both are reactants that participate in the reaction.
When they mix in solution, they form a precipitate of silver chloride
  • (AgCl)
  • and sodium nitrate (NaNO鈧�).
This is represented by the balanced chemical equation:

\[\text{AgNO}_3 \text{ (aq)} + \text{NaCl (aq)} \rightarrow \text{AgCl (s)} + \text{NaNO}_3 \text{ (aq)}\]
This equation indicates that one molecule of
  • AgNO鈧� reacts with one molecule of NaCl
  • to produce one molecule of AgCl
  • and one molecule of NaNO鈧�.
Writing chemical equations helps us understand the quantity and relationship of substances involved in the reactions.
Molar Mass
Molar mass is a crucial concept in stoichiometry as it connects the mass of a substance to the amount in moles. It is the mass of one mole of a given substance, often expressed in grams per mole (g/mol).
In solving this exercise, you need to know the molar mass of sodium chloride (NaCl) to find out how much mass is needed.
The molar mass of NaCl is the sum of the molar masses of its constituent atoms:
  • The molar mass of sodium (Na) is approximately 22.99 g/mol,
  • and that of chlorine (Cl) is approximately 35.45 g/mol.
Adding these gives the molar mass of NaCl as 58.44 g/mol.
Knowing the molar mass allows us to convert from moles to grams, helping us find that approximately 0.6574 grams of NaCl are needed.
Precipitation Reactions
Precipitation reactions occur when two aqueous solutions mix and an insoluble solid forms. This solid is called a precipitate. In this problem, the precipitation reaction occurs between the silver nitrate (础驳狈翱鈧�) and sodium chloride (NaCl) solutions.
When AgNO鈧� is added to NaCl, silver chloride (AgCl) precipitates as a solid. This happens because Ag鈦� ions in silver nitrate combine with Cl鈦� ions in sodium chloride.
  • The result is the formation of a white solid, AgCl, which is insoluble in water.
  • The remaining sodium ions (Na鈦�) and nitrate ions (NO鈧冣伝) stay dissolved in the solution.
Precipitation reactions help in removing ions from solutions, useful in various chemical processes especially analytical chemistry to identify specific ions.
Solution Concentration
Solution concentration describes how much solute is present in a given volume of solvent, often expressed in moles per liter (Molarity, M). In this exercise, we start with a silver nitrate solution of 0.2500 M concentration in a 45.0 mL volume.
Understanding molarity is key to determining how much
  • NaCl is needed to react with the AgNO鈧� solution.
To calculate the amount of Ag鈦� ions present, convert the volume from mL to L, and then multiply by the concentration:
  • Volume = 0.045 L (since 1 L = 1000 mL)
  • Moles = Volume 脳 Concentration = 0.045 L 脳 0.2500 M = 0.01125 mol of Ag鈦�
The stoichiometry of the reaction tells us equal moles of NaCl are needed to completely precipitate the Ag鈦� ions, leading us to calculate the mass of NaCl required based on the number of moles.

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Most popular questions from this chapter

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+} ?\)

Ignoring protolysis reactions (i.e. proton transfer reaction), specify what ions are present in a solution upon dissolving each of the following substances in water: (a) \(\mathrm{Li}_{2} \mathrm{CO}_{3},\) (b) \((\mathrm{NH} 4)_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{NaPF}_{6^{\circ}}\) (c) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+& \\ & 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{MH}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

The concentration of alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) in blood, called the "blood alcohol concentration" or BAC, is given in units of grams of alcohol per \(100 \mathrm{~mL}\) of blood. The legal definition of intoxication, in many states of the United States, is that the BAC is 0.08 or higher. What is the concentration of alcohol, in terms of molarity, in blood if the BAC is \(0.08 ?\)

An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of \(\mathrm{NaCl}\) of the same concentration. Which of the following substances could the unknown be: \(\mathrm{KOH}\), \(\mathrm{NH}_{3}, \mathrm{HNO}_{3}, \mathrm{KClO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{3}, \mathrm{CH}_{3} \mathrm{COCH}_{3}\) (acetone)?
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