Which salt is the least soluble acoorinding to equilibrium constant?

[匿名]

2016-08-05 7:34 am

59) The solubility of 1:1 salts is measured by the equilibrium constant for the general reaction: MX(s) = Mn+(aq) + Xn-(aq). Given the following salts and their equilibrium constants for the reaction above at 25°C, which salt is the least soluble?
A) MgCO3, Kc = 6.8 × 10-6
B) CaCO3, Kc = 5.0 × 10-9
C) SrCO3, Kc = 5.6 × 10-10
D) BaCO3, Kc = 2.6 × 10-9

60) Salt solubilities can be compared by the concentration of cation formed when the salt dissolves in the general reaction: MaXb(s) ⇌ a Mb+(aq) + b Xa-(aq). Given the following salts and their equilibrium constants for the reaction above at 25°C, which salt is the least soluble?
A) AgCl, Kc = 1.8 × 10-10
B) Ag2SO4, Kc = 1.2 × 10-5
C) CaCO3, Kc = 2.6 × 10-9
D) CaF2, Kc = 1.5 × 10-10

Please explain how to answer the questions, thanks!

回答 (1)

不用客氣

2016-08-05 8:28 am

✔ 最佳答案

59)
All of the 4 salts are 1 : 1 salts. Therefore, the smaller the Kc (Ksp), the less soluble the salt is.
SrCO₃ is the least soluble because it has the smallest Kc.

Answer: SrCO₃, Kc = 5.6 × 10⁻¹⁰

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60)
Let s mol/L be the solubility of the salt MaXb.
MaXb(s) ⇌ aMᵇ⁺(aq) + bXᵃ⁻(aq)
At equilibrium : [Mᵇ⁺] = as mol/L and [Xᵃ⁻] = bs mol/L
Kc = [Mᵇ⁺][Xᵃ⁻]
Hence, Kc = (as)ᵃ (bs)ᵇ

In A, AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
s × s = 1.8 × 10⁻¹⁰
Then, s = 1.3 × 10⁻⁵ (the smallest s)

In B, Ag₂SO₄(s) ⇌ 2Ag⁺(aq) + SO₄²⁻(aq)
(2s)² × s = 1.2 × 10⁻⁵
Then, s = 1.4 × 10⁻²

In C, CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq)
s × s = 2.6 × 10⁻⁹
Then, s = 5.1 × 10⁻⁵

In D, CaF₂(s) ⇌ Ca⁺(aq) + 2F⁻(aq)
s × (2s)² = 1.5 × 10⁻¹⁰
Then, s = 3.3 × 10⁻⁴

Answer: A) AgCl, Kc = 1.8 × 10⁻¹⁰

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Alternative solution to 60)

Both A and C involve 1 : 1 salts.
Kc(AgCl) < Kc(CaCO₃)
Hence, solubility: AgCl < CaCO₃

B involves a 2 : 1 salt, while D involves a 1 : 2 salt.
Kc(CaF₂) < Kc(Ag₂SO₄)
Hence, solubility: CaF₂ < Ag₂SO₄

Kc of AgCl is close to Kc of CaF₂.
AgCl is a 1 : 1 salt, while CaF₂ is a 1 : 2 salt.
Hence, solubility : AgCl < CaF₂

Answer: A) AgCl, Kc = 1.8 × 10⁻¹⁰

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