Question

Using the Gibbs energy change, ∆G° = +63.3 kJ,
for the following reaction,
Ag, CO, = 2Ag+ (aq) + CO3 2- (aq)
the Ksp of Ag2CO3(s) in water at 25°C is :-
(R = 8.314 J K-1 mol-1)
(1) 3.2 x 10-26 (2) 8.0 x 10-12
(3) 2.9 x 10-3 (4) 7.9 x 10-2 WITH FULL SOLUTION!​

Answer 1

Answer: (2)   $8.0\times 10^{-12}$

Explanation:

Formula used :

$Ag_2CO_3\rightleftharpoons 2Ag^+(aq)+CO_3^{2-}(aq)$

$\Delta G^o=-2.303\times RT\times \log K_sp$

where,

$\Delta G^0$ = standard Gibbs free energy change = 63.3 kJ = 63300 J

R = universal gas constant = 8.314 J/K/mole

T = temperature = $25^0C=(25+273)=298 K$

$K_sp$ = soklubility product = ?

$63300 J=-2.303\times 8.314\times \times 298\log K_sp$

$K_sp=8.0\times 10^{-12}$

Thus solubility product of $Ag_2CO_3$ is $8.0\times 10^{-12}$