/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 131 The concentration of lead ions \... [FREE SOLUTION] | 91Ó°ÊÓ

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The concentration of lead ions \(\left(\mathrm{Pb}^{2+}\right)\) in a sample of polluted water that also contains nitrate ions \(\left(\mathrm{NO}_{3}^{-}\right)\) is determined by adding solid sodium sulfate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right)\) to exactly \(500 \mathrm{~mL}\) of the water. (a) Write the molecular and net ionic equations for the reaction. (b) Calculate the molar concentration of \(\mathrm{Pb}^{2+}\) if \(0.00450 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). was needed for the complete precipitation of \(\mathrm{Pb}^{2+}\) ions as \(\mathrm{PbSO}_{4}\).

Short Answer

Expert verified
The molecular equation is \(Na_2SO_4 + Pb^{2+} \rightarrow PbSO_4 + 2Na^+\) and the net ionic equation is \(\mathrm{SO}_{4}^{2-} + Pb^{2+} \rightarrow PbSO_4(s)\). The molar concentration of \(Pb^{2+}\) in the solution is 0.0634 mM.

Step by step solution

01

Write the molecular equation

Sodium sulfate (\(Na_2SO_4\)) reacts with lead ions (\(Pb^{2+}\)) to form lead sulfate (\(PbSO_4\)) and Sodium ions (\(Na^+\)). The reaction will look like this: \(Na_2SO_4 + Pb^{2+} \rightarrow PbSO_4(s) + 2Na^+\). Sodium ions are spectactor ions.
02

Write the net ionic equation

\(\mathrm{SO}_{4}^{2-} + Pb^{2+} \rightarrow PbSO_4(s)\). In the net ionic equation, you focus on the ions actively involved in the reaction, so you leave out the Sodium ions.
03

Calculate the number of moles of \(Na_2SO_4\)

To calculate the number of moles, use the formula: moles = mass / molar mass. The molar mass of \(Na_2SO_4\) is approximately 142 g/mol. So, moles = 0.00450 g / 142 g/mol = 0.0000317 mol.
04

Concentration of \(Pb^{2+}\) in the Solution

The stoichiometry of the reaction shows that one mole of \(Na_2SO_4\) reacts with one mole of \(Pb^{2+}\). Meaning, the number of moles of \(Pb^{2+}\) in the system is also 0.0000317 mol. Concentration of \(Pb^{2+}\) in the solution is calculated by dividing the number of moles of \(Pb^{2+}\) by the volume of the solution in liters. So, concentration = 0.0000317 moles / 0.5 L = 0.0000634 mol/L or 0.0634 mM.

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Key Concepts

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

Molecular and Net Ionic Equations
Chemical reactions can be represented through different types of equations, with molecular equations showing the complete formulas of all reactants and products, while net ionic equations only display the species that actually participate in the reaction.

For instance, when sodium sulfate reacts with lead ions to form lead sulfate in water, the molecular equation would include all the involved compounds, including spectator ions which do not directly participate in the reaction. The complete molecular equation for our exercise is:
\[Na_2SO_4 + Pb^{2+} \rightarrow PbSO_4(s) + 2 Na^+\]
In contrast, the net ionic equation simplifies this by showing just the chemical species that undergo a change during the reaction. In this case, the sodium ions (\(Na^+\)) are spectator ions and are not included in the net ionic equation. The resulting net ionic equation for the precipitation reaction is:
\[SO_4^{2-} + Pb^{2+} \rightarrow PbSO_4(s)\]
Understanding both types of equations is pivotal in studying and predicting the outcome of chemical reactions. It helps in identifying the actual reactants and the formation of precipitates, crucial for processes like water treatment.
Molar Concentration Calculation
The concentration of a solution is a measure of the amount of solute present per unit volume of solution. It is commonly expressed in molarity (M), which is the number of moles of solute per liter of solution.

To calculate molarity, you can use the formula:
\[Molarity = \frac{number\ of\ moles\ of\ solute}{volume\ of\ solution\ in\ liters}\]
In our exercise, the molar concentration of lead ions (\(\text{Pb}^{2+}\)) is determined by calculating the moles of \(\text{Pb}^{2+}\) from the mass of sodium sulfate that reacts and subsequently dividing by the volume of the solution where the reaction took place. It's important to ensure that the mass of the solute is converted into moles using the solute's molar mass, and that the solution's volume is in liters for this calculation.
Stoichiometry of a Reaction
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using the balanced chemical equation to determine the proportions of reactants needed to form a product.

In stoichiometric calculations, it is essential to first convert known quantities (usually mass or volume) to moles, using the molar mass for solids and liquids or molar volume for gases at standard temperature and pressure (STP). In the provided exercise, the stoichiometry indicates a 1:1 mole ratio between sodium sulfate and lead ions. It implies that each mole of \(Na_2SO_4\) reacts completely with one mole of \(\text{Pb}^{2+}\).

After finding the moles of the precipitant, the mole ratio is used to calculate the moles of the targeted ion in solution. Finally, to find the molar concentration of that ion, divide the number of moles by the volume of the solution. This stoichiometric approach is a cornerstone in chemistry helping in the analysis of reaction mixtures, optimization of product yields, and the calculation of reactants required for industrial processes.

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Most popular questions from this chapter

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+& 6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q) \end{aligned}\) Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of oxalic acid.

Describe in each case how you would separate the cations or anions in an aqueous solution of: (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) (b) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{KNO}_{3}\), (c) \(\mathrm{KBr}\) and \(\mathrm{KNO}_{3}\), (d) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KNO}_{3}\), (e) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaNO}_{3}\).

Which of the following aqueous solutions would you expect to be the best conductor of electricity at \(25^{\circ} \mathrm{C} ?\) Explain your answer. (a) \(0.20 \mathrm{M} \mathrm{NaCl}\) (b) \(0.60 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (c) \(0.25 M \mathrm{HCl}\) (d) \(0.20 M \operatorname{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

You are given a colorless liquid. Describe three chemical tests you would perform on the liquid to show that it is water.

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