The Ksp value for magnesium carbonate, MgCO3, is 7.2 x 10-6. If 2.50 g of magnesium carbonate is placed in 1.00 x 102 mL of water, how much (in g) magnesium carbonate will dissolve?
Well, if I had to guess, I'd say that magnesium carbonate would probably be a bit shy and not want to dissolve too much. It's like the introverted kid at a party who would rather hang out by themselves. But anyway, let's see if we can figure this out.
The first thing we need to do is convert the volume of water from milliliters to liters, because the Ksp value is typically given in terms of molarity. So, 1.00 x 10² mL is equal to 0.1 L.
Now, to find out how much magnesium carbonate will dissolve, we can use the Ksp value and some stoichiometry. The molar mass of MgCO3 is approximately 84 g/mol.
If x grams of magnesium carbonate dissolve, we can write the expression for Ksp as:
Ksp = [Mg²⁺][CO3²⁻]
Since MgCO3 ionizes into Mg²⁺ and CO3²⁻, the concentration of Mg²⁺ and CO3²⁻ will be equal. Let's call the concentration of dissolved magnesium carbonate "s".
Ksp = s²
Now, we can substitute the values into the equation:
7.2 x 10⁻⁶ = s²
Taking the square root of both sides gives us:
s ≈ 8.5 x 10⁻⁴ M
To find the amount of magnesium carbonate that dissolves in grams, we can calculate the moles of MgCO3 using its molar mass:
moles = concentration × volume
moles = 8.5 x 10⁻⁴ mol/L × 0.1 L
moles ≈ 8.5 x 10⁻⁵ mol
Finally, we can convert moles to grams:
grams = moles × molar mass
grams = 8.5 x 10⁻⁵ mol × 84 g/mol
grams ≈ 7.1 x 10⁻³ g
So, approximately 7.1 x 10⁻³ grams (or 0.0071 g) of magnesium carbonate will dissolve in 2.50 g placed in 1.00 x 10² mL of water. But remember, this is just an approximation, as real-world factors can influence the solubility.
To find how much magnesium carbonate will dissolve in water, we need to compare the Q value (reaction quotient) with the Ksp value.
Step 1: Write the balanced equation for the dissolution of magnesium carbonate:
MgCO3(s) ⇌ Mg2+(aq) + CO3^2-(aq)
Step 2: Calculate the molar mass of MgCO3:
Molar mass of MgCO3 = (24.31 g/mol) + (12.01 g/mol) + (3 × 16.00 g/mol)
= 24.31 g/mol + 12.01 g/mol + 48.00 g/mol
= 84.32 g/mol
Step 3: Convert the given mass of magnesium carbonate to moles:
Number of moles of MgCO3 = (2.50 g) / (84.32 g/mol)
≈ 0.0297 mol
Step 4: Convert the given volume of water to liters:
Volume of water = (1.00 x 10^2 mL) / (1000 mL/L)
= 1.00 L
Step 5: Calculate the initial concentration of MgCO3:
Initial concentration of MgCO3 = (0.0297 mol) / (1.00 L)
≈ 0.0297 M
Step 6: Calculate the Q value:
Q = [Mg2+][CO3^2-]
Since MgCO3 has a 1:1 stoichiometry, the concentration of Mg2+ and CO3^2- will be equal. Therefore,
Q = [Mg2+][CO3^2-] = (0.0297 M)(0.0297 M) = 0.000881 M^2
Step 7: Compare the Q value with the Ksp value:
Ksp = 7.2 x 10^-6
Since the Q value (0.000881 M^2) is larger than the Ksp value (7.2 x 10^-6), we have an excess of MgCO3 in solution. Therefore, all of the MgCO3 will dissolve.
Step 8: Calculate the mass of magnesium carbonate that will dissolve:
Mass of magnesium carbonate = (0.0297 mol) × (84.32 g/mol)
= 2.50 g
Therefore, the amount of magnesium carbonate that will dissolve in this case is 2.50 grams.
To determine the amount of magnesium carbonate that will dissolve in water, we can use the solubility product constant (Ksp) and the given mass and volume values.
The Ksp expression for magnesium carbonate is as follows:
Ksp = [Mg2+][CO32-]
We know that magnesium carbonate dissociates as follows:
MgCO3(s) ⇌ Mg2+(aq) + CO32-(aq)
Let's assume that x grams of magnesium carbonate dissolve in water. Since 2.50 g of magnesium carbonate is initially placed in water, the expression for the dissolved magnesium carbonate can be represented as follows:
[Mg2+] = x g / (2.50 g)
Given that the volume of water is 1.00 x 102 mL, we can convert it to liters:
Volume = 1.00 x 102 mL × (1 L / 1000 mL) = 1.00 L
So, we can determine the concentration of Mg2+ and CO32- ions in terms of moles per liter:
[Mg2+] = x g / (2.50 g) = (x / 2.50) mol/L
[CO32-] = (x / 2.50) mol/L
Now, substitute these values into the Ksp expression:
Ksp = (x / 2.50) × (x / 2.50) = (x^2) / (2.50 * 2.50)
Let's solve for x:
(x^2) = Ksp * (2.50 * 2.50)
x^2 = 7.2 x 10^(-6) * 6.25
x^2 = 4.5 x 10^(-5)
x = √(4.5 x 10^(-5))
x ≈ 0.0067 mol
Now, we can calculate the mass of magnesium carbonate (in grams) that will dissolve:
Mass = x mol * (84.31 g / mol)
Mass ≈ 0.0067 mol * 84.31 g/mol
Mass ≈ 0.564 g
Therefore, approximately 0.564 grams of magnesium carbonate will dissolve in 1.00 x 10^2 mL of water.