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

[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

answer: h+ its less than oh

explanation:

yes

explanation:

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