Solubility Products

To determine the molar solubility of zirconium phosphate, let's first understand the dissolution process of zirconium phosphate, , in water.

The dissolution of zirconium phosphate can be represented by the following equilibrium equation:

Given the complexity of zirconium often being highly hydrolysable and not forming simple ion complexes easily, let's focus on the simplified attribute: each mole of zirconium phosphate dissolves to give one mole of ions and one mole of ions into the solution.

The equilibrium constant for this reaction, often referred to as the solubility product constant , is expressed in terms of molar solubility as follows:

At equilibrium:

and

Thus,

To solve for the molar solubility in terms of , we take the square root of both sides:

<.p>From the given options, the correct expression for the molar solubility of zirconium phosphate matches the option:

To determine the order of precipitation when NH₄OH is added to a solution containing 1M A²⁺ and 1M B³⁺ ions, we need to calculate the concentration of hydroxide ions [OH^-] required to reach the solubility product (K_sp) of each hydroxide.

1. First, consider the dissociation of A(OH)₂:

The solubility product expression is:

Given and assuming equilibrium concentrations are approached as precipitation starts, with , we find:

Solve for [OH^-]:

1. Now, consider the dissociation of B(OH)₃:

The solubility product expression is:

Given and with , we find:

Solve for [OH^-]:

1. Comparison: The concentration of OH^- needed for B(OH)₃ to precipitate is , which is lower than the required for A(OH)₂. This indicates B(OH)₃ will precipitate first as its K_sp is reached at a lower [OH^-] concentration.

Conclusion: B(OH)₃ will precipitate before A(OH)₂.

To find the molar solubility of in water, we begin by understanding the dissolution equation and the relationship between solubility product ( ) and molar solubility ( ).

The dissolution equilibrium of in water can be given by:

The expression for the solubility product is:

If is the molar solubility of , then the concentration of will be and the concentration of will be . Substituting these into the expression for :

Given , we substitute this value into the equation:

Solving for , we get:

Taking the fourth root of both sides gives:

Thus, the molar solubility of in water is given by the expression: