1.Sodium hydroxide is slowly added to a solution that is 0.067 M in Cu2+ and 0.1

Question

1.Sodium hydroxide is slowly added to a solution that is 0.067 M in Cu2+ and 0.138 M in Fe3+.    (Ksp = 2.2 x 1020 for copper (II) hydroxide; Ksp = 1.1 x 1036 for iron (III) hydroxide)
a.Which compound will precipitate first?
b.Calculate the concentration of the first cation precipitated that still remains in solution when the second cation starts to precipitate. 1.Sodium hydroxide is slowly added to a solution that is 0.067 M in Cu2+ and 0.138 M in Fe3+.    (Ksp = 2.2 x 1020 for copper (II) hydroxide; Ksp = 1.1 x 1036 for iron (III) hydroxide)
a.Which compound will precipitate first?
b.Calculate the concentration of the first cation precipitated that still remains in solution when the second cation starts to precipitate. 1.Sodium hydroxide is slowly added to a solution that is 0.067 M in Cu2+ and 0.138 M in Fe3+.    (Ksp = 2.2 x 1020 for copper (II) hydroxide; Ksp = 1.1 x 1036 for iron (III) hydroxide)
a.Which compound will precipitate first?
b.Calculate the concentration of the first cation precipitated that still remains in solution when the second cation starts to precipitate.

Explanation / Answer

1. a.

Only the Fe3+ ions will get precipitated as Fe(OH)3, and the Ca2+ ions will almost entirely remain in solution.

An ionic compound is precipitated from solution when the product of the concentrations of its cation and anion exceeds its solubility product.

Since the solubility product of Fe(OH)3 is very low, it gets precipitated even with such a low concentration of OH- ions. The solubility product of Ca(OH)2 is much higher. So, the Ca2+ ions will not get precipitated.

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