The Ksp of manganese (II) hydroxide, Mn(OH)2 is 2.00x10^-13. Calculate the solubility of this compound in g/L?
See your post on Ag2CrO4.Worked the same way with solubility in M. Convert that to grams.
To calculate the solubility of manganese (II) hydroxide (Mn(OH)2) in grams per liter (g/L), you need to follow these steps:
Step 1: Write the balanced chemical equation. The solubility of a compound can be expressed by the equation: Mn(OH)2 ↔ Mn^2+ + 2OH^-
Step 2: Determine the moles of Mn(OH)2 dissolved. Let's assume the solubility of Mn(OH)2 is 'x' moles per liter.
Step 3: Use the equation to determine the concentration of Mn^2+ ions in terms of 'x': [Mn^2+] = x moles/liter.
Step 4: Since Mn(OH)2 dissociates into one Mn^2+ ion, the concentration of Mn(OH)2 will also be 'x' moles/liter.
Step 5: Use the equation to determine the concentration of OH^- ions in terms of 'x': [OH^-] = 2x moles/liter.
Step 6: Since Mn(OH)2 dissociates into two OH^- ions, the concentration of OH^- ions in Mn(OH)2 will be 2x moles/liter.
Step 7: Use the solubility-product constant, Ksp, expression for Mn(OH)2, which is: Ksp = [Mn^2+][OH^-]^2.
Step 8: Substitute the concentrations calculated in steps 4 and 6 into the Ksp expression: Ksp = (x)(2x)^2 = 4x^3.
Step 9: Substitute the given value of Ksp (2.00x10^-13) into the equation from step 8: 2.00x10^-13 = 4x^3.
Step 10: Solve the equation obtained in step 9 for 'x'. Take the cube root of both sides: x = (2.00x10^-13/4)^(1/3).
Step 11: Calculate the value of 'x' using the equation from step 10. The value obtained for 'x' is the solubility of Mn(OH)2 in moles per liter.
Step 12: Convert the solubility of Mn(OH)2 from moles per liter to grams per liter. To do this, multiply 'x' by the molar mass of Mn(OH)2.
Therefore, by following these steps, you can calculate the solubility of manganese (II) hydroxide (Mn(OH)2) in grams per liter (g/L).