Determine the concentrations of Na2CO3, Na , and CO32– in a solution prepared by dissolving 2.57 × 10–4 g Na2CO3 in 1.50 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of CO32–. Ignore any reactions with water
mols Na2CO3 = grams/molar mass = ?
Then M Na2CO3 = mols/L (but the problem SHOULD SAY in 1.50 L of SOLUTION).
(Na^+) = 2 x (Na2CO3)
(CO3^2-) = (Na2CO3)
For ppm, convert 2.57E-4 g Na2CO3 in 1.50 L to mg/L and that will be the ppm.
To determine the concentrations of Na2CO3, Na, and CO32- in the solution, we will first calculate the molarity and then convert it to parts per million (ppm).
1. Calculate the molar mass of Na2CO3:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of Na2CO3 = (2 × Na) + C + (3 × O) = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
2. Calculate the number of moles of Na2CO3:
Moles = Mass (g) / Molar mass (g/mol)
Moles = 2.57 × 10^(-4) g / 105.99 g/mol ≈ 2.42 × 10^(-6) mol
3. Calculate the molarity (M) of Na2CO3:
Molarity = Moles / Volume (L)
Molarity = 2.42 × 10^(-6) mol / 1.50 L ≈ 1.61 × 10^(-6) M
4. Convert the molarity to parts per million (ppm):
ppm = Molarity × 10^6
ppm = 1.61 × 10^(-6) M × 10^6 = 1.61 ppm (approximately)
5. Since Na2CO3 dissociates into two Na+ ions and one CO32- ion, the molar concentration of Na+ is twice the molar concentration of Na2CO3 (1.61 × 2 = 3.22 ppm).
6. Similarly, the molar concentration of CO32- is the same 1.61 ppm.
Therefore, the concentrations of Na2CO3, Na, and CO32- in the solution are as follows:
- Na2CO3 concentration: 1.61 × 10^(-6) M (or 1.61 ppm)
- Na+ concentration: 3.22 × 10^(-6) M (or 3.22 ppm)
- CO32- concentration: 1.61 × 10^(-6) M (or 1.61 ppm)
To determine the concentrations of Na2CO3, Na, and CO32– in the given solution, we need to follow these steps:
Step 1: Calculate the molar mass of Na2CO3.
The molar mass of Na2CO3 is the sum of the molar masses of its constituent atoms:
Molar mass of Na2CO3 = (2 * atomic mass of Na) + (1 * atomic mass of C) + (3 * atomic mass of O)
Using the atomic masses from the periodic table:
Molar mass of Na2CO3 = (2 * 22.99 g/mol) + (1 * 12.01 g/mol) + (3 * 16.00 g/mol)
Molar mass of Na2CO3 = 105.99 g/mol
Step 2: Calculate the moles of Na2CO3.
We can use the given mass of Na2CO3 and its molar mass to calculate the moles:
moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
moles of Na2CO3 = 2.57 × 10–4 g / 105.99 g/mol
Step 3: Calculate the molarity of Na2CO3.
Molarity is defined as moles of solute divided by liters of solution.
molarity of Na2CO3 = moles of Na2CO3 / volume of solution
volume of solution = 1.50 L
molarity of Na2CO3 = moles of Na2CO3 / volume of solution
Step 4: Calculate the concentrations in parts per million (ppm).
Converting molarity to parts per million (ppm) involves multiplying by a factor of 10^6.
concentration in ppm = molarity x 10^6
Step 5: Repeat steps 3 and 4 for Na and CO32–.
Since Na2CO3 dissociates into two Na+ ions and one CO32– ion, we need to divide the molarity and concentration by 2 for Na and multiply by 2 for CO32–.
Following these steps, you should be able to determine the concentrations of Na2CO3, Na, and CO32– in the given solution expressed in molarity and parts per million (ppm).