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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

Using the Gibbs energy change, ° = + 63.3kJ , for the following...

Using the Gibbs energy change, \DeltaG°=+63.3kJ\DeltaG ° = + 63.3\text{kJ} , for the following reaction,
Ag2CO3→2Ag+(aq)+CO32−(aq)A{g}_{2}C{O}_{3} \to 2Ag + ( aq ) + C{O}_{3}^{2 -} ( aq )
the KspofAg2CO3(s){K}_{sp} of A{g}_{2}C{O}_{3} ( s ) in water at 25oC{25}^{o}C is
(R=8.314JK−1mol−1R = 8.314\text{J}{\text{K}}^{- 1}mo{l}^{- 1})\left\{\text{Given:}{10}^{0.91} ≅ 8\right\}

Options

  1. A7.9×10−27.9 \times 1{0}^{- 2}
  2. B8.0×10−128.0 \times 1{0}^{- \text{12}}
  3. C2.9×10−32.9 \times 1{0}^{- 3}
  4. D3.2×10−263.2 \times 1{0}^{- \text{26}}

Correct answer

B. 8.0×10−128.0 \times 1{0}^{- \text{12}}

Step-by-step solution

{\text{\Delta G}}^{\text{o}} = - 2.303\text{RT}\text\log{\text{K}}_{\text{sp}}
63.3×1000=−2.303×8.314×298logKsp63.3 \times 1000 = - 2.303 \times 8.314 \times 298log {K}_{sp}
logKsp=−11.09log {K}_{sp} = - 11.09
Ksp=10−11.09=8×10−12{K}_{sp} = 1{0}^{- 11.09} = 8 \times 1{0}^{- 12}
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