Question

A chemist prepares a solution by dissolving 55.0 g of aluminum nitrate in 1250 ml of water.

a) write the dissociation equation for an aqueous solution of aluminum nitrate. that is, write the balanced equation to show the dissolving of the solute to produce ions.
b) calculate the molar concentration of each ion in the solution.

Answer 1

The dissociation equation for aluminum nitrate in water is Al(NO3)3(s) → Al3+(aq) + 3NO3−(aq). The molar concentrations of the ions are obtained by dividing the moles of aluminum nitrate by the volume of the solution in liters and multiplying by three for the nitrate ions.

Step-by-step explanation:

The student asked about the dissolution of aluminum nitrate in water and the calculation of molar concentration of the resulting ions. First part asks to write the dissociation equation for aluminum nitrate in water, and the second part involves calculating the molar concentration of each ion from the given mass and volume.

Part A: Dissociation Equation

Aluminum nitrate dissociates in water according to the equation:

Al(NO3)3(s) → Al3+(aq) + 3NO3−(aq)

Part B: Molar Concentration of Ions

First, we convert the mass of aluminum nitrate to moles using its molar mass. Then we use the volume of water to find the molarity of aluminum ions. Since every mole of aluminum nitrate produces three moles of nitrate ions, their molarity will be three times that of the aluminum ions.

Example: If 55.0 g of aluminum nitrate (Al(NO3)3) are dissolved in 1250 mL of water, the calculation would follow these steps:

1. Calculate moles of Al(NO3)3: Moles = Mass (g) / Molar Mass (g/mol).
2. Calculate the molarity of Al3+: Molarity = Moles of Al(NO3)3 / Volume (L).
3. Calculate the molarity of NO3−: Molarity = 3 * Molarity of Al3+ since there are 3 nitrate ions for every aluminum ion.