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Chapter 14: Problem 93

Calculate the mass of AgCl formed, and the concentration of silver ion remaining in solution, when \(10.0 \mathrm{g}\) of solid \(\mathrm{AgNO}_{3}\) is added to \(50 . \mathrm{mL}\) of \(1.0 \times\) \(10^{-2} M\) NaCl solution. Assume there is no volume change upon addition of the solid.

Short Answer

Expert verified
The mass of AgCl formed is 0.716 g, and the remaining concentration of silver ion in the solution is 1.078 M.

Step by step solution

01

Write the balanced chemical equation

We first need to write the balanced chemical equation for the reaction between AgNO鈧� and NaCl: \[ AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq) \]
02

Calculate the molar amount of AgNO鈧� and NaCl

We are given that 10.0 g of AgNO鈧� are added to 50 mL of a \(1.0\times10^{-2} \,M\) NaCl solution. First, we'll find the moles of both the given reagents. For AgNO鈧�, the molecular weight is: 1 Ag = 107.87 g/mol 1 N = 14.01 g/mol 3 O = 3 * 16.00 g/mol = 48.00 g/mol 1 AgNO鈧� = 107.87 + 14.01 + 48.00 g/mol = 169.88 g/mol Moles of AgNO鈧� = (10.0 g) / (169.88 g/mol) = 0.0589 mol For NaCl, Moles of NaCl = (50 mL) * (1.0 * 10鈦宦� mol/L) = 0.005 mol
03

Determine the limiting reagent and the amount of AgCl formed

Comparing the molar amount of AgNO鈧� and NaCl: 0.0589 mol AgNO鈧� (1 mol AgCl/1 mol AgNO鈧�) = 0.0589 mol AgCl 0.005 mol NaCl (1 mol AgCl/1 mol NaCl) = 0.005 mol AgCl Since 0.005 mol AgCl < 0.0589 mol AgCl, NaCl is the limiting reagent. There will be 0.005 moles of AgCl formed in the solution. To find the mass of AgCl formed, we need its molecular weight: 1 Ag = 107.87 g/mol 1 Cl = 35.45 g/mol 1 AgCl = 107.87 + 35.45 g/mol = 143.32 g/mol Mass of AgCl formed = (0.005 mol) * (143.32 g/mol) = 0.716 g
04

Calculate the remaining concentration of silver ion in the solution using Ksp

Now that we know the amount of AgCl precipitated, we can find the remaining Ag鈦� ion concentration in the solution. First, calculate the remaining moles of AgNO鈧�: Moles of remaining AgNO鈧� = moles of initial AgNO鈧� - moles of AgCl formed = 0.0589 mol - 0.005 mol = 0.0539 mol As the volume of the solution doesn't change upon the addition of solid, we can calculate the concentration of Ag鈦� in the solution: Concentration of Ag鈦� = moles of remaining AgNO鈧�/volume of solution = 0.0539 mol/(50.0 mL + 0 mL) * (1 L/1000 mL) = 1.078 \,M The remaining concentration of silver ion in the solution is 1.078 M. To summarize: Mass of AgCl formed = 0.716 g Remaining concentration of silver ion in the solution = 1.078 M

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Key Concepts

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

Chemical Reaction Equation
Understanding the chemical reaction equation is crucial for stoichiometry problem solving. It represents the process wherein reactants transform into products. Here, we have the equation:
\[ AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq) \]
The letters in the parentheses identify the state of each compound: (aq) for aqueous solutions and (s) for solids. By balancing this equation, we ensure that the number of atoms for each element is the same on both sides. This reflects the Law of Conservation of Mass, stating that matter is neither created nor destroyed in a chemical reaction.
Limiting Reagent
The concept of the limiting reagent is a fundamental aspect of stoichiometry. It's the reactant that will be completely consumed first, thus determining the end of the chemical reaction and the maximum amount of product that can form. To find it, we compare the mole ratio of the reactants with the coefficients in the balanced equation. In our exercise, we identify that NaCl is the limiting reagent because the moles of NaCl are less than the moles of AgNO鈧� when compared to the stoichiometric ratio from the balanced chemical equation. Therefore, it dictates the amount of precipitate, AgCl, that will form.
Molarity
Molarity is a way to express concentration, denoted as M, which is defined as moles of solute divided by liters of solution. In stoichiometry problems, it allows us to determine the amount of reactants in a given volume of solution. For instance, in the exercise, the initial molarity of NaCl is given to be \(1.0 \times 10^{-2} \, M\), which helps us find the moles of NaCl in solution. After the reaction, we calculate the molarity of the remaining Ag鈦� to assess how much reactant is left unreacted.
Precipitate Formation
Precipitate formation occurs when two aqueous solutions combine to form an insoluble solid. This solid is what we refer to as a precipitate. The formation of AgCl as a white solid precipitate from the reaction between AgNO鈧� and NaCl is an example of this process. To calculate the mass of the precipitate formed, we first need to identify the limiting reagent, because it restricts the total amount of product that can be formed. In our case study, we calculated that exactly 0.716 grams of AgCl will precipitate out of the solution.

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Most popular questions from this chapter

For each of the following, the mass of the solute is given, followed by the total volume of the solution prepared. Calculate the molarity. a. \(1.25 \mathrm{g} \mathrm{KNO}_{3} ; 115 \mathrm{mL}\) b. \(12.5 \mathrm{g} \mathrm{KNO}_{3} ; 1.15 \mathrm{L}\) c. \(1.25 \mathrm{mg} \mathrm{KNO}_{3} ; 1.15 \mathrm{mL}\) d. \(1.25 \mathrm{kg} \mathrm{KNO}_{3} ; 115 \mathrm{L}\)

How much water must be added to \(500 .\) mL of 0.200 \(M \mathrm{HCl}\) to produce a \(0.150 \mathrm{M}\) solution? (Assume that the volumes are additive.)

How many milliliters of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}\) solution are required to precipitate all the nickel, as NiS, from 25.0 \(\mathrm{mL}\) of \(0.20 \mathrm{M} \mathrm{NiCl}_{2}\) solution? \(\mathrm{NiCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{S}(a q) \rightarrow \mathrm{NiS}(s)+2 \mathrm{NaCl}(a q)\)

One way to determine the amount of chloride ion in a water sample is to titrate the sample with standard \(\mathrm{AgNO}_{3}\) solution to produce solid AgCl. $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{AgCl}(s)$$ If a 25.0 -mL water sample requires \(27.2 \mathrm{mL}\) of \(0.104 M \mathrm{AgNO}_{3}\) in such a titration, what is the concentration of \(\mathrm{Cl}^{-}\) in the sample?

The total acidity in water samples can be determined by neutralization with standard sodium hydroxide solution. What is the total concentration of hydrogen ion, \(\mathrm{H}^{+},\) present in a water sample if \(100 .\) mL of the sample requires \(7.2 \mathrm{mL}\) of \(2.5 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\) to be neutralized?
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