How many milligrams of magnesium palmitate will precipitate from 965ml of this solution when it is cooled from 50C to 25C? For Mg(C16H21O2)2, Ksp= 4.8x10(-12) at 50C and 3.3x10(-12) at 25C.
Mg(pal)2 ==> Mg^2+ + 2pal^-
....x.........x.......2x
Ksp = (Mg^2+)(pal)^2/[Mg(pal)2]
Substitute into Ksp expression and use Ksp @ 50C to determine the solubility at 50 C. x = solubility in moles/L. Convert to grams/L, then to grams/965 mL.
Repeat the above procedure using Ksp @ 25 C and determine grams/965 mL.
Then subtract grams/965mL of the two to determine the grams that will ppt and convert to mg.
Ksp = (Mg^2+)(pal)^2/[Mg(pal)2]- expression is incorrect. mg(pal)2 is not part of the ksp expression as the reactant is in the solid state. only Ksp = [Mg^2+][pal]^2 is relevant.
thank you!
To calculate the number of milligrams of magnesium palmitate that will precipitate from the given solution, we need to use the solubility product constant (Ksp) and the temperature change.
First, let's determine the molar mass of magnesium palmitate (Mg(C16H21O2)2):
- Molar mass of magnesium (Mg) = 24.31 g/mol
- Molar mass of palmitate (C16H21O2) = (12.01 g/mol x 16) + (1.01 g/mol x 21) + (16.00 g/mol x 2) = 278.41 g/mol
The molar mass of magnesium palmitate is the sum of the molar masses of magnesium and two palmitate molecules:
Molar mass of Mg(C16H21O2)2 = 24.31 g/mol + 2 * 278.41 g/mol = 581.13 g/mol
Now, let's calculate the concentration of magnesium palmitate in the given solution:
- Volume of solution = 965 ml = 0.965 L
- Concentration (in moles per liter) = molar mass / volume = 581.13 g/mol / 0.965 L ≈ 602.07 g/L
To find the number of milligrams of magnesium palmitate that will precipitate, we need to calculate the concentration when the solution is cooled from 50°C to 25°C using the solubility product constant (Ksp):
- At 50°C, Ksp = 4.8x10^(-12)
- At 25°C, Ksp = 3.3x10^(-12)
The solubility of magnesium palmitate decreases with a decrease in temperature. The Ksp values provide a ratio of ion concentrations in saturated solutions. We can set up the following equation:
Ksp = [Mg²⁺][C16H21O2]²
Let's assume x represents the concentration (in moles per liter) of magnesium palmitate that will precipitate at 25°C. Therefore, the concentration of magnesium ions ([Mg²⁺]) will also be x, and the concentration of palmitate ions ([C16H21O2]) will be 2x.
Using the given Ksp values:
At 50°C: 4.8x10^(-12) = (x)(2x)² = 4x³
At 25°C: 3.3x10^(-12) = (x)(2x)² = 4x³
Now, let's solve these equations to find the value of x, which represents the concentration (in moles per liter) of magnesium palmitate that will precipitate at 25°C.
At 50°C:
4x³ = 4.8x10^(-12)
Dividing both sides by 4: x³ = 1.2x10^(-12)
Taking the cube root: x = (1.2x10^(-12))^(1/3) ≈ 5.53x10^(-5) M
At 25°C:
4x³ = 3.3x10^(-12)
Dividing both sides by 4: x³ = 8.25x10^(-13)
Taking the cube root: x = (8.25x10^(-13))^(1/3) ≈ 4.47x10^(-5) M
Now, let's calculate the amount of magnesium palmitate that will precipitate at 25°C using the concentration and volume of the given solution:
- Mass (in grams) = concentration x volume
- Mass (in milligrams) = (concentration x volume) x 1000
Mass of magnesium palmitate precipitated = (4.47x10^(-5) M) x (0.965 L) x (581.13 g/mol) x (1000 mg/g)
= 24.76 mg
Therefore, approximately 24.76 milligrams of magnesium palmitate will precipitate from the 965 ml solution when it is cooled from 50°C to 25°C.