Determine the concentrations of MgCl2, Mg2+, and Cl- in a solution prepared by dissolving 2.39 x 10^-4g
MgCl2 in 2.50L of water. Express all three concentrations in molarity. Also display the concentrations of ionic species in part per million (ppm).
To determine the concentrations of MgCl2, Mg2+, and Cl- in the solution, we first need to calculate the number of moles of MgCl2 dissolved in the water.
Step 1: Convert the mass of MgCl2 to moles.
Given: Mass of MgCl2 = 2.39 x 10^-4 g.
The molar mass of MgCl2 can be calculated from the periodic table:
Mg = 24.31 g/mol
Cl = 35.45 g/mol (x2) = 70.9 g/mol
MgCl2 = 24.31 + 70.9 = 95.21 g/mol
To convert the mass to moles, we'll use the formula:
moles = mass / molar mass
moles of MgCl2 = 2.39 x 10^-4 g / 95.21 g/mol
Step 2: Calculate the concentration of MgCl2 in molarity.
Molarity (M) is defined as moles of solute divided by volume of solution in liters (L):
Molarity (M) = moles / volume
Given: Volume = 2.50 L
Concentration of MgCl2 in molarity = moles of MgCl2 / volume
Step 3: Calculate the concentration of Mg2+ and Cl-.
In 1 mole of MgCl2, there is 1 mole of Mg2+ ions and 2 moles of Cl- ions.
Concentration of Mg2+ in molarity = concentration of MgCl2 in molarity
Concentration of Cl- in molarity = 2 x concentration of MgCl2 in molarity
Step 4: Calculate the concentrations in parts per million (ppm).
To convert the concentration from molarity to ppm, use the conversion factor:
1 ppm = 1 mg/L = 1 mg/kg = 1 mg/g = 1 g/10^6 g
Concentration in ppm = concentration in molarity x molar mass (g/mol) x 10^6
Now you have the step-by-step procedure to determine the concentrations of MgCl2, Mg2+, and Cl-. You can substitute the given values into the equations to find the answers.
To determine the concentrations of MgCl2, Mg2+, and Cl- in the solution, we need to follow these steps:
Step 1: Calculate the moles of MgCl2.
Given mass of MgCl2 = 2.39 x 10^-4 g
Molar mass of MgCl2 = 95.211 g/mol
Moles of MgCl2 = (2.39 x 10^-4 g) / (95.211 g/mol) = 2.51 x 10^-6 mol
Step 2: Calculate the concentration of MgCl2.
Concentration of MgCl2 = Moles of MgCl2 / Volume of solution
Concentration of MgCl2 = (2.51 x 10^-6 mol) / (2.50 L) = 1.004 x 10^-6 M
Step 3: Calculate the concentrations of Mg2+ and Cl- ions.
Since MgCl2 dissociates fully in water, the concentration of Mg2+ ions is equal to the concentration of MgCl2.
Therefore, the concentration of Mg2+ = 1.004 x 10^-6 M
The concentration of Cl- ions is twice the concentration of MgCl2 since each MgCl2 molecule dissociates into one Mg2+ ion and two Cl- ions.
Therefore, the concentration of Cl- = 2 * 1.004 x 10^-6 M = 2.008 x 10^-6 M
Step 4: Calculate the concentrations in parts per million (ppm).
Concentration in ppm = (Concentration in M) * (Molar mass) * 10^6
Concentration of MgCl2 in ppm = (1.004 x 10^-6 M) * (95.211 g/mol) * 10^6 = 95.572 ppm
Concentration of Mg2+ in ppm = (1.004 x 10^-6 M) * (24.305 g/mol) * 10^6 = 24.573 ppm
Concentration of Cl- in ppm = (2.008 x 10^-6 M) * (35.453 g/mol) * 10^6 = 71.399 ppm
So, the concentrations in molarity are:
- MgCl2: 1.004 x 10^-6 M
- Mg2+: 1.004 x 10^-6 M
- Cl-: 2.008 x 10^-6 M
And the concentrations in parts per million (ppm) are:
- MgCl2: 95.572 ppm
- Mg2+: 24.573 ppm
- Cl-: 71.399 ppm
mols MgCl2 = grams/molar mass = ?
M MgCl2 = mols MgCl2/L = ?
Technically, there isn't any MgCl2 since all of it is ionized completely; however, I think the spirit of the question is to give the concentration of the salt, MgCl2.
You will have 2.39E-4 g Mg ^2+ there and that is 0.239 mg Mg^2+ in 2.5 L. How much is that in mg/L? That's 0.239 mg/2.5 L = approx (that's estimated) 0.096 mg/L. Since 1 mg/L = 1 ppm that is 0approx 0.096 ppm for Mg^+ and twice that for Cl^-.