Standard solutions of sodium carbonate, Na2CO3(aq), are required for certain types of chemical analyses. If 8.50 g ofsodium carbonate is dissolved in 500 mL of distilled water, calculate the concentration of the sodium ions and the concentration of the carbonate ions dissolved in this solution. Hint: In order to solve this problem, write the correctly balanced dissociation equation.
Standard solutions of sodium carbonate, Na2CO3(aq), are required for certain types of chemical
analyses.
If 8.50 g ofsodium carbonate is dissolved in 500 mL of distilled water, calculate the concentration
of the sodium ions and the concentration of the carbonate ions dissolved in this solution.ÂÂ
Hint: In order to solve this problem, write the correctly balanced dissociation equation.
The Correct Answer and Explanation is :
To calculate the concentration of sodium ions and carbonate ions in the solution, we need to follow these steps:
- Write the Dissociation Equation: Sodium carbonate (Na₂CO₃) dissociates in water according to the following equation:
[
\text{Na}_2\text{CO}_3 (s) \rightarrow 2\text{Na}^+ (aq) + \text{CO}_3^{2-} (aq)
]
This equation tells us that each formula unit of sodium carbonate produces two sodium ions and one carbonate ion. - Calculate the Molar Mass of Sodium Carbonate: The molar mass of Na₂CO₃ is calculated by summing the molar masses of its constituent elements.
- Sodium (Na): (22.99 \, \text{g/mol})
- Carbon (C): (12.01 \, \text{g/mol})
- Oxygen (O): (16.00 \, \text{g/mol}) per atom and there are three oxygen atoms.
[
\text{Molar mass of Na}_2\text{CO}_3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 \, \text{g/mol}
]
- Calculate Moles of Sodium Carbonate Dissolved:
[
\text{Moles of Na}_2\text{CO}_3 = \frac{8.50 \, \text{g}}{105.99 \, \text{g/mol}} \approx 0.0802 \, \text{mol}
] - Calculate Moles of Ions Produced: From the stoichiometry of the dissociation equation, each mole of Na₂CO₃ produces two moles of Na⁺ and one mole of CO₃²⁻.
- Moles of Na⁺: (2 \times 0.0802 \, \text{mol} = 0.1604 \, \text{mol})
- Moles of CO₃²⁻: (0.0802 \, \text{mol})
- Calculate the Concentrations:
- Volume of solution = 500 mL = 0.500 L
- Concentration of Na⁺ ions:
[
[\text{Na}^+] = \frac{0.1604 \, \text{mol}}{0.500 \, \text{L}} = 0.3208 \, \text{M}
] - Concentration of CO₃²⁻ ions:
[
[\text{CO}_3^{2-}] = \frac{0.0802 \, \text{mol}}{0.500 \, \text{L}} = 0.1604 \, \text{M}
]
This calculation demonstrates the process of determining the concentration of ions in a solution from a known quantity of a solid salt. Understanding the stoichiometry of the salt’s dissociation in water is critical. Each mole of sodium carbonate provides specific amounts of sodium and carbonate ions, directly affecting the ionic concentration in the solution.
For the image, let’s create a visual that illustrates the dissociation of sodium carbonate in water, showing the formula units breaking into ions and their respective concentrations.
Here is a visual illustration of the dissociation of sodium carbonate (Na₂CO₃) in water, showing how the compound splits into sodium ions (Na⁺) and carbonate ions (CO₃²⁻), along with their calculated concentrations in the solution. This image can help in visualizing the chemical process and understanding the concept better.
