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Calculate the molar solubility of ag2so4 in each solution below. the ksp of silver sulfate is 1.5 × 10−5. (a) 0.36 m agno3: (b) 0.36 m na2so4:

Respuesta :

superman1987

Answer:

a) 1.157 x 10⁻⁴ M.

b) 0.0032 M.

Explanation:

  • Ag₂SO₄ is sparingly soluble salt in water which is dissociate according to:

Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻.

Ksp = [Ag⁺][SO₄²⁻] = (2s)²(s) = 1.5 x 10⁻⁵.

4s³ = 1.5 x 10⁻⁵.

s³ = 1.5 x 10⁻⁵/4 = 3.75 x 10⁻⁶.

∴ s = ∛(3.75 x 10⁻⁶) = 0.0155 M.

a) 0.36 M AgNO₃:

  • The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of AgNO₃:

AgNO₃ → Ag⁺ + NO₃⁻.

  • So, the concentration of Ag⁺ will be that of AgNO₃  and Ag₂SO₄:

[Ag⁺] = (2s + 0.36)², s is neglected with respect to 0.36 M the concentration resulted from AgNO₃.

So, [Ag⁺] = (0.36)².

∴ Ksp = [Ag⁺][SO₄²⁻] = (0.36)²(s) = 1.5 x 10⁻⁵.

∴ s = 1.5 x 10⁻⁵/(0.36)² = 1.157 x 10⁻⁴ M.

b) 0.36 M Na₂SO₄:

  • The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of Na₂SO₄:

Na₂SO₄ → 2Na⁺ + SO₄²⁻.

  • So, the concentration of SO₄²⁻ will be that of Na₂SO₄  and Ag₂SO₄:

[SO₄²⁻] = (s + 0.36), s is neglected with respect to 0.36 M the concentration resulted from Na₂SO₄.

So, [SO₄²⁻] = (0.36).

∴ Ksp = [Ag⁺][SO₄²⁻] = (2s)²(0.36) = 1.5 x 10⁻⁵.

∴ 4s² = 1.5 x 10⁻⁵/(0.36) = 4.166 x 10⁻⁵.

∴ s² = 4.166 x 10⁻⁵/4 = 1.042 x 10⁻⁵.

∴ s = √(1.042 x 10⁻⁵) = 0.0032 M.

dsdrajlin

The molar solubility of Ag₂SO₄ in 0.36 M AgNO₃ is 1.2 × 10⁻⁴ M. The molar solubility of Ag₂SO₄ in 0.36 M Na₂SO₄ is 3.2 × 10⁻³ M.

We want to calculate the molar solubility of Ag₂SO₄ (S) in different solutions. We will make ICE charts, considering the common ion provided by the other salt.

(a) 0.36 M AgNO₃

The initial concentration of Ag⁺ is 0.36 M.

         Ag₂SO₄(s) ⇄ 2 Ag⁺ + SO₄²⁻(aq)

I                                0.36      0

C                               +2S     +S

E                          0.36+2S     S

The solubility product constant (Ksp) is:

[tex]Ksp = [Ag^{+}]^{2} [SO_4^{2-} ] = (0.36+2S)^{2} S\approx 0.36^{2} S\\\\S = \frac{Ksp}{0.36^{2} } = \frac{1.5 \times 10^{-5} }{0.36^{2} } = 1.2 \times 10^{-4} M[/tex]

(b) 0.36 M Na₂SO₄

The initial concentration of SO₄²⁻ is 0.36 M.

         Ag₂SO₄(s) ⇄ 2 Ag⁺ + SO₄²⁻(aq)

I                                    0       0.36

C                               +2S        +S

E                                 2S       0.36 + S

The solubility product constant (Ksp) is:

[tex]Ksp = [Ag^{+}]^{2} [SO_4^{2-} ] = (2S)^{2} (0.36+S) \approx 4S^{2} (0.36)\\\\S = \sqrt{\frac{Ksp}{1.44} } = \sqrt{\frac{1.5 \times 10^{-5} }{1.44} } = 3.2 \times 10^{-3}M[/tex]

The molar solubility of Ag₂SO₄ in 0.36 M AgNO₃ is 1.2 × 10⁻⁴ M. The molar solubility of Ag₂SO₄ in 0.36 M Na₂SO₄ is 3.2 × 10⁻³ M.

Learn more: https://brainly.com/question/1776962

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