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Using the square-bracket notation, express the concentration of ions in a solution that is 0.811 M in aluminum sulfate [Al2(SO4)3].

[Al^3+] = ____ M

[SO4^2−] = ____ M

Respuesta :

dsdrajlin

Answer:

[Al³⁺] = 1.62 M

[SO₄²⁻] = 2.43 M

Explanation:

Step 1: Given data

Molar concentration of Al₂(SO₄)₃: 0.811 M

Step 2: Write the reaction for the ionization of Al₂(SO₄)₃

Al₂(SO₄)₃(aq) ⇒ 2 Al³⁺(aq) + 3 SO₄²⁻(aq)

Step 3: Calculate the molar concentration of Al³⁺

The molar ratio of Al₂(SO₄)₃ to Al³⁺ is 1:2. The molarity of Al³⁺ is 2/1 × 0.811 M = 1.62 M.

Step 4: Calculate the molar concentration of SO₄²⁻

The molar ratio of Al₂(SO₄)₃ to SO₄²⁻ is 1:3. The molarity of SO₄²⁻ is 3/1 × 0.811 M = 2.43 M

pstnonsonjoku

The concentration of the aluminium ion is 0.4055 M while the concentration of the sulfate ion is 0.270  M.

We know that the following reaction occurs for the dissolution of Al2(SO4)3 in water;

Al2(SO4)3(s) ----> 2Al^3+(aq) + 3SO4^2-(aq)

To find the concentration of each of the ions;

  • Concentration of Al^3+ = 1/2 × 0.811 M = 0.4055 M

So using the square bracket notation to indicate concentration;

[Al^3+] = 0.4055 M

  • Concentration of SO4^2- = 1/3 × 0.811 M = 0.270 M

So using the square bracket notation to indicate concentration;

[SO4^2−] =0.270  M

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