Chemistry 30

Write balanced reaction equation that show which ions are produced when the following substances are dissolved in water.

Solution:

Iron(III) nitrate has a solubility of 0.15 M. Find concentration of the ions in solution.

Solution:

Begin by writing a balanced dissociation equation:

Fe(NO₃)₃ → Fe³⁺_(aq) + 3 NO₃^-_(aq)

The concentration of the ions can be determined from the balancing coefficients from the equation:

[Fe³⁺] = 1× [Fe(NO₃)₃] = 1 × 0.15 = 0.15 M

[NO₃^-] = 3× [Fe(NO₃)₃] = 3 × 0.15 = 0.45 M

Calculate ion concentrations in a 2.00 L solution containing 17.1 g aluminum sulfate, Al₂(SO₄)₃

Solution:

Before calculating the concentration of the ions, we must first calculate the concentration of the aluminum sulfate solution.

We will need to find the molar mass of Al₂(SO₄)₃:

Calculate the concentration of Al₂(SO₄)₃:

Write a balanced equation for the dissociation reaction:

Al₂(SO₄)₃ → 2 Al³⁺_(aq) + 3 SO₄^2-_(aq)

Using the balanced equation, calculate the concentration of the individual ions:

[Al³⁺] = 2× [Al₂(SO₄)₃] = 2 × 0.0249 = 0.0498 M or 4.98 ×10^-2M

[SO₄^2-] = 3× [Al₂(SO₄)₃] = 3 × 0.0249 = 0.0747 M or 7.47×10^-2M