Problem 44 - Chemistry: The Central Science

Question

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} \mathrm{M} .\) The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

Step-by-Step Solution

Answer

In conclusion, to reduce the concentration of \(\mathrm{Ca}^{2+}\) ions to 20% of its original level in 5.0 × \(10^{7} \mathrm{~L}\) of water, \(4.223 \times 10^{6}\) g of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(4.505 \times 10^{6}\) g of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) must be added.

Step 1: Calculate the moles of \(\mathrm{Ca}^{2+}\) ions in the water supply

To determine the moles of \(\mathrm{Ca}^{2+}\) ions in the water supply, we can use the formula: Moles of \(\mathrm{Ca}^{2+}\) = Concentration of \(\mathrm{Ca}^{2+}\) x Volume of water Moles of \(\mathrm{Ca}^{2+}\) = \(5.7 \times 10^{-3} \mathrm{M} * 5.0 \times 10^{7} \mathrm{L} = 2.85 \times 10^{5} \) moles

Step 2: Calculate the moles of \(\mathrm{Ca}^{2+}\) ions remaining after reducing concentration

To find the moles of \(\mathrm{Ca}^{2+}\) ions remaining after reducing the concentration to 20% of its original level, we can multiply the initial moles by 20%: Moles of remaining \(\mathrm{Ca}^{2+}\) = \(2.85 \times 10^{5} \mathrm{~moles} * 0.2 = 5.7 \times 10^{4} \mathrm{~moles}\)

Step 3: Calculate the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed

From the balanced chemical reaction, we know: \(\mathrm{Ca^{2+}} + 2\mathrm{OH^{-}} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}\) Since the reaction ratio is 1:1, the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed will be equal to the moles of remaining \(\mathrm{Ca}^{2+}\) ions: Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed = 5.7 × \(10^{4} \mathrm{~moles}\)

Step 4: Calculate the mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed

Now we will calculate the mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed by multiplying the moles by its molar mass: Mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed = Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) x Molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) Molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) = 40.08 (Ca) + 2(15.999) + 2(1.008) = 74.093 g/mol Mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed = 5.7 × \(10^{4}\) moles × 74.093 g/mol = \(4.223 \times 10^{6}\) g

Step 5: Calculate the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed

From the balanced chemical reaction, we know: 2\(\mathrm{HCO_3^{-}} +\mathrm{Na}_{2}\mathrm{CO_{3}} \rightarrow 2\mathrm{CO_{2}} + 2\mathrm{H}_{2}\mathrm{O} + 2\mathrm{Na}^+ + 2\mathrm{OH^-}\) Since the reaction ratio is 2:1, the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed will be half the moles of \(\mathrm{HCO}_{3}^-\) ions: Moles of \(\mathrm{HCO}_{3}^{-}\) ions = \(1.7 \times 10^{-3} \mathrm{M} * 5.0 \times 10^{7} \mathrm{L} = 8.5 \times 10^{4} \mathrm{~moles}\) Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed = (8.5 × \(10^{4} \mathrm{~moles}) / 2 = 4.25 × \(10^{4} \mathrm{~moles}\)

Step 6: Calculate the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed

Now we will calculate the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed by multiplying the moles by its molar mass: Mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed = Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) × Molar mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) Molar mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) = 2(22.989) + 12.01 + 3(15.999) = 105.988 g/mol Mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed = 4.25 × \(10^{4}\) moles × 105.988 g/mol = \(4.505 \times 10^{6}\) g In conclusion, the necessary masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) to be added are \(4.223 \times 10^{6}\) g and \(4.505 \times 10^{6}\) g, respectively.

Key Concepts

Molar Concentration

Molar concentration, often referred to as molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This is a crucial concept in chemistry, especially when dealing with solutions and reactions.

• The molar concentration formula is given by: \[ ext{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
• This allows us to express how concentrated a solution is.

Understanding molarity is vital for preparing solutions with accurate concentrations for chemical reactions. For example, in the provided exercise, the molarity of calcium ions \(\mathrm{Ca}^{2+}\) and bicarbonate ions \(\mathrm{HCO}_{3}^{-}\) in the water is found to determine how much additional compound is required. By using the molar concentration and the volume, we are able to calculate the total number of moles present and adjust concentrations as needed.

Stoichiometry Calculations

Stoichiometry is the part of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is like a recipe that guides us in understanding how much of each ingredient is needed, ensuring that reactions proceed as expected.

To perform stoichiometry calculations:

• First, balance the chemical equation involved. This means ensuring that you have the same number of each type of atom on both sides of the reaction.
• Use the coefficients in the balanced equation to set up molar ratios that relate reactants to products.
• Convert known quantities (like masses or volumes) to moles using molar masses.
• Use these molar ratios and known moles to calculate unknown quantities.

In the exercise, stoichiometry allows us to determine the exact masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed to achieve a desired chemical balance in water. Accurate stoichiometry ensures complete reaction and avoids excess of any chemicals in the solution.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. These reactions are represented by chemical equations that show how reactants change into products.

• A balanced chemical equation is essential. It ensures the law of conservation of mass is respected, meaning atoms are neither created nor destroyed.
• Reactions can be classified into several types, such as synthesis, decomposition, single replacement, double replacement, and combustion.

In the exercise problem, understanding the reactions helps in manipulating the concentrations of ions in water. The addition of \(\mathrm{Ca}(\mathrm{OH})_{2}\) reduces \(\mathrm{Ca}^{2+}\) ions, while \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) reacts to adjust bicarbonate levels. By altering these concentrations, one can control and maintain the desired chemical equilibrium within the water, demonstrating the dynamic nature and importance of chemical reactions in practical applications.