Problem 72 - Chemistry: An Atoms-Focused Approach

Question

Given the following equation, how many grams of \(\mathrm{Pb} \mathrm{CO}_{3}\) will dissolve when 1.00 L of \(1.00 M \mathrm{H}^{+}\) is added to \(5.00 \mathrm{g}\) of \(\mathrm{PbCO}_{3} ?\) $$ \mathrm{PbCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(g) $$

Step-by-Step Solution

Answer

Solution: \(5.00\, \text{g}\) of \(\mathrm{Pb} \mathrm{CO}_{3}\) will dissolve in the solution.

Step 1: Calculate the number of moles of \(\mathrm{H}^{+}\)

In the problem, we are given that there are 1.00 L of \(1.00 M \mathrm{H}^{+}\). The formula to calculate moles is: $$ \text{Moles} = \text{Volume} \times \text{Molarity} $$ So the number of moles of \(\mathrm{H}^{+}\) is: $$ \text{Moles}(\mathrm{H}^{+}) = 1.00\,\text{L} \times 1.00\, \text{M} = 1.00\, \text{mol} $$

Step 2: Determine the stoichiometry of \(\mathrm{Pb}\mathrm{CO}_{3}\) to \(\mathrm{H}^{+}\) in the equation

In the given balanced equation, we can see that for every 2 moles of \(\mathrm{H}^{+}\), 1 mole of \(\mathrm{Pb} \mathrm{CO}_{3}\) is dissolved. $$ \mathrm{PbCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(g) $$

Step 3: Calculate the number of moles of \(\mathrm{Pb} \mathrm{CO}_{3}\) that dissolve

Knowing the stoichiometry from Step 2, we can calculate the number of moles of \(\mathrm{Pb} \mathrm{CO}_{3}\) that dissolve when 1.00 moles of \(\mathrm{H}^{+}\) are added to the solution. $$ \text{Moles}(\mathrm{PbCO}_3) = \frac{1}{2} \times \text{Moles}(\mathrm{H}^{+}) = \frac{1}{2} \times 1.00\, \text{mol} = 0.50\, \text{mol} $$

Step 4: Convert the moles of \(\mathrm{Pb} \mathrm{CO}_{3}\) back to grams

Now, we need to convert the moles of \(\mathrm{Pb} \mathrm{CO}_{3}\) back to grams. The molar mass of \(\mathrm{Pb} \mathrm{CO}_{3}\) is approximately \(267.2\, \text{g}/\text{mol}\). The formula to convert moles to grams is: $$ \text{Grams} = \text{Moles} \times \text{Molar Mass} $$ So the grams of \(\mathrm{Pb} \mathrm{CO}_{3}\) that dissolve are: $$ \text{Grams}(\mathrm{PbCO}_3) = 0.50\, \text{mol} \times 267.2\, \text{g}/\text{mol} = 133.6\, \text{g} $$ Since we initially have \(5.00\, \text{g}\) of \(\mathrm{Pb} \mathrm{CO}_{3}\), all of it will dissolve when \(1.00\, \text{L}\) of \(1.00\, \text{M}\, \mathrm{H}^{+}\) is added. Thus, \(5.00\, \text{g}\) of \(\mathrm{Pb} \mathrm{CO}_{3}\) will dissolve in this reaction.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that provides the quantitative relationship between reactants and products in a chemical reaction. In this exercise, we use stoichiometry to understand the dissolution process of \( \mathrm{PbCO}_{3} \) in the presence of \( \mathrm{H}^{+} \). It helps us determine how many moles of reactants are needed or will be used, and how many moles of products will be formed.

In the given chemical equation:

• 1 mole of \( \mathrm{PbCO}_{3}(s) \) reacts with 2 moles of \( \mathrm{H}^{+}(aq) \).
• This relationship is crucial as it guides us to work out the amount of \( \mathrm{PbCO}_{3} \) that will dissolve.

By knowing that 2 moles of \( \mathrm{H}^{+} \) dissolve 1 mole of \( \mathrm{PbCO}_{3} \), we calculate that if we have 1 mole of \( \mathrm{H}^{+} \), then 0.5 moles of \( \mathrm{PbCO}_{3} \) will dissolve.

This basic stoichiometric relationship is foundational to predict and balance reactions, ensuring the law of conservation of mass is obeyed in chemical processes.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds, resulting in new products. The exercise provided focuses on a reaction between \( \mathrm{PbCO}_{3} \) and \( \mathrm{H}^{+} \) ions, producing \( \mathrm{Pb}^{2+} \), \( \mathrm{H}_{2} O \), and \( \mathrm{CO}_{2} \).

This type of reaction highlights an important principle: solubility can be influenced by chemical reactions. Here, the reaction helps dissolve solid \( \mathrm{PbCO}_{3} \) by effectively removing \( \mathrm{CO}_{3}^{2-} \) ions as \( \mathrm{CO}_{2} \) gas. This shifts the equilibrium, favoring the dissolution of more \( \mathrm{PbCO}_{3} \) into ions.

• Solid \( \mathrm{PbCO}_{3} \) interacts with aqueous \( \mathrm{H}^{+} \) to form gaseous \( \mathrm{CO}_{2} \), a key factor that drives the reaction.
• The presence of water (\( \mathrm{H}_{2}O \)) as a product reassures that the aqueous environment facilitates the reaction.

Understanding how chemical reactions affect solubility can aid in predicting how different compounds will behave in solution, which is profoundly important in fields like environmental science and pharmaceuticals.

Molar Mass

Molar mass is critical in converting between the mass of a substance and the amount in moles, a conversion required to relate physical quantities in chemical formulas.

• For \( \mathrm{PbCO}_{3} \), a crucial task in this exercise involves calculating its molar mass to move between grams and moles effectively.
• The molar mass is calculated from the atomic masses of lead (Pb), carbon (C), and oxygen (O).

Explicitly, the calculation is as follows:

• \( \mathrm{Pb} \): approximately 207.2 g/mol
• \( \mathrm{C} \): approximately 12.01 g/mol
• \( \mathrm{O} \): 3 atoms with approximately 16.00 g/mol each

Summing these values gives us \( 267.2 \) g/mol for \( \mathrm{PbCO}_{3} \).

Using molar mass to convert moles into grams lets us determine how many grams of \( \mathrm{PbCO}_{3} \) can dissolve: \( 0.50 \, \text{mol} \times 267.2 \, \text{g/mol} = 133.6 \, \text{g} \).

Comprehending molar mass helps link microscopic particle amounts to measurable, real-world quantities, making it a cornerstone of chemical calculations.