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Chapter 4: Problem 56

What volume of \(0.100 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from \(150.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) ?

Short Answer

Expert verified
\(250.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{3}\mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from \(150.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

Step by step solution

01

Write the balanced chemical equation

First, we write the balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2: \[2\mathrm{Na}_{3}\mathrm{PO}_{4}(aq) + 3\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(aq) \rightarrow 6\mathrm{NaNO}_{3}(aq) + \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\]
02

Calculate the moles of Pb(NO3)2

Next, we calculate the moles of Pb(NO3)2 by multiplying its concentration by its volume: Moles of Pb(NO3)2 = concentration × volume Moles of Pb(NO3)2 = 0.250 M × 150.0 mL × \(\frac{1L}{1000mL}\) = 0.0375 mol
03

Use stoichiometry to determine the moles of Na3PO4 required

Now, we will use the stoichiometry of the balanced equation to calculate the moles of Na3PO4 required to precipitate all the lead(II) ions: From the balanced equation, for every 3 moles of Pb(NO3)2, 2 moles of Na3PO4 are required. So, Moles of Na3PO4 = \(\frac{2 \,mol \,\mathrm{Na}_{3}\mathrm{PO}_{4}}{3\, mol \,\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}\) × 0.0375 mol of Pb(NO3)2 = 0.025 mol
04

Calculate the volume of Na3PO4 required

Finally, we will use the molarity of Na3PO4 to calculate the volume of the solution required for the reaction: Volume of Na3PO4 = Moles of Na3PO4 / Concentration of Na3PO4 Volume of Na3PO4 = 0.025 mol / 0.100 M = 0.250 L or 250.0 mL So, 250.0 mL of 0.100 M Na3PO4 is required to precipitate all the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2.

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Key Concepts

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

Balancing Chemical Equations
Understanding chemical equations is crucial in chemistry. A chemical equation represents a chemical reaction where the reactants are shown on the left side and the products on the right. To obey the Law of Conservation of Mass, the equation must be balanced, meaning it should have the same number of atoms of each element on both sides.

Take, for example, the reaction of sodium phosphate \textbf{(Na}\(_3\)\textbf{PO}\(_4\)\textbf{)} with lead(II) nitrate \textbf{(Pb(NO}\(_3\)\textbf{)}\(_2\)\textbf{)}. The equation is balanced by adjusting the coefficients to ensure the same number of atoms of each element appears in both reactants and products. This process can be challenging, but it provides critical information for stoichiometric calculations. A balanced equation is a proportion that allows us to predict how much of each substance is involved in a reaction.
Molar Concentration Calculations
The concept of molarity, symbolized as 'M', is a way of expressing the concentration of a solution. It is measured in moles per liter, indicating how many moles of a solute are present in one liter of solution. In the given exercise, the molarity of lead(II) nitrate is given as 0.250 M.

To calculate the amount of lead(II) nitrate in moles, we need to multiply its molar concentration by the volume of the solution, which must be in liters. By performing these calculations correctly, we obtain the moles of lead(II) nitrate, which are essential for the subsequent stoichiometric calculations. This step is fundamental to determine how much of another substance will react with or be produced from a given amount of the solute.
Stoichiometric Calculations
Stoichiometry is the mathematical analysis of the quantitative relationships, or ratios, defined by a balanced chemical equation. It allows us to predict the amounts of products and reactants that are consumed and produced in a chemical reaction.

In the exercise, the stoichiometric ratio between lead(II) nitrate and sodium phosphate is used to find out how much sodium phosphate is needed to completely react with the lead(II) nitrate present. This involves the use of the mole ratio, derived from the balanced equation, which in this case shows that two moles of sodium phosphate react with three moles of lead(II) nitrate. With the mole ratio and the moles of lead(II) nitrate calculated from the molar concentration, we can prescribe the precise amount of sodium phosphate required. This is a core concept in stoichiometry, highlighting its role as the bridge between chemical identities and their quantities in reactions.

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Most popular questions from this chapter

Consider an experiment in which two burets, \(\mathrm{Y}\) and \(\mathrm{Z}\), are simultaneously draining into a beaker that initially contained \(275.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{HCl}\). Buret \(\mathrm{Y}\) contains \(0.150 \mathrm{M} \mathrm{NaOH}\) and buret \(Z\) contains \(0.250 \mathrm{M} \mathrm{KOH}\). The stoichiometric point in the titration is reached \(60.65\) minutes after \(\bar{Y}\) and \(Z\) were started simultaneously. The total volume in the beaker at the stoichiometric point is \(655 \mathrm{~mL}\). Calculate the flow rates of burets \(\mathrm{Y}\) and \(\mathrm{Z}\). Assume the flow rates remain constant during the experiment.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 121 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

A \(10.00-\mathrm{mL}\) sample of sulfuric acid from an automobile battery requires \(35.08 \mathrm{~mL}\) of \(2.12 M\) sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.

Specify which of the following are oxidation-reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical transformers. Because PCBs have been shown to be potentially harmful, analysis for their presence in the environment has become very important. PCBs are manufactured according to the following generic reaction: $$ \mathrm{C}_{12} \mathrm{H}_{10}+{ }_{n \mathrm{Cl}_{2}} \rightarrow \mathrm{C}_{12} \mathrm{H}_{10-n} \mathrm{Cl}_{n}+n \mathrm{HCl} $$ This reaction results in a mixture of \(\mathrm{PCB}\) products. The mixture is analyzed by decomposing the \(\mathrm{PCBs}\) and then precipitating the resulting \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). a. Develop a general equation that relates the average value of \(n\) to the mass of a given mixture of \(\mathrm{PCBs}\) and the mass of \(\mathrm{AgCl}\) produced. b. A \(0.1947-g\) sample of a commercial PCB yielded \(0.4791 \mathrm{~g}\) of \(\mathrm{AgCl}\). What is the average value of \(n\) for this sample?
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