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Using the Gibb’s energy change,ΔG⁰ = +63.3 kJ, for the following reaction ,Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq) the Ksp of Ag₂CO₃(s) in water at 25⁰C is
Options
(a) 3.2 ˣ 10⁻²⁶
(b) 8.0 ˣ 10⁻¹²
(c) 2.9 ˣ 10⁻³
(d) 7.9 ˣ 10⁻²
Correct Answer:
8.0 ˣ 10⁻¹²
Explanation:
∆G°=-2.303RT log KSP
here K=[Ag⁺]² [CO₃⁻⁻] =K (sp)
.·. 63.3×10⁻³=-2.303×8.314 ×298 log K(sp)
.·. log K(sp) = -63.3×10⁻³/5705.8 =-11.09
.·. K(sp) = Antilog(-11.09) = 8×10⁻¹².
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Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
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