A 600ml sample of a saturated solution of MgCO3 is reduced to 125 ml by evaporation. What mass of MgCo3 is formed? Ksp=4*10^-5
MgCO3 ==>Mg^2+ + CO3^2-
Ksp = (Mg^2+)(CO3^2-)
If you let x = solublity MgCO3, then x = (Mg^&2+) and x = (CO3^2-). Substitute and solve for x = solubility MgCO3.
That will be mols/L. Convert to mols in 600 mL and call this y1 mols to start.
Then convert to mols in 125 mL and call this y2 mols. The difference between y2 and y1 will be what ppts. Convert that to grams.
thank you!
To find the mass of MgCO3 formed, we need to use the information given and make use of the solubility product constant (Ksp).
Here's how we can proceed:
1. Calculate the initial concentration of MgCO3 in the saturated solution using the given volume:
Initial concentration (C1) = Volume (V1) = 600 mL
2. Calculate the final concentration of MgCO3 after evaporation using the reduced volume:
Final concentration (C2) = Volume (V2) = 125 mL
3. Since MgCO3 is a sparingly soluble salt, we assume that it completely dissociates into Mg2+ and CO3^2- ions when it dissolves. Therefore, the concentration of both ions is the same as the concentration of MgCO3.
4. Use the formula for Ksp:
Ksp = [Mg2+][CO3^2-]
5. Since the concentration of Mg2+ ions is the same as the concentration of MgCO3, we can substitute:
Ksp = [MgCO3] * [MgCO3]
6. Rearrange the equation to solve for [MgCO3]:
[MgCO3] = sqrt(Ksp)
7. Calculate [MgCO3] using the given Ksp:
[MgCO3] = sqrt(4 * 10^-5)
8. Now multiply the concentration by the final volume to find the number of moles (n):
n = [MgCO3] * V2
9. Convert the moles of MgCO3 to grams by multiplying by its molar mass:
Mass = n * molar mass of MgCO3
By following the steps above, you will be able to find the mass of MgCO3 formed.