Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium
I 2 (s) 
I – (aq.) + 
(aq.)
When the equilibrium concentrations at 300 K are [I – ] = 0.10 M and [ 
] = 0.10 M.
{Given → 
(I – , aq.) = – 50 KJ/mol,
( 
, aq.) = – 123.5 KJ/mol,
(H 2 O, λ ) = – 233 KJ/mol
(OH – , aq.) = – 150 KJ/mol
R (Gas constant) = 
J/mol–K
Log e = 2.3}
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
8
Sol. Balanced equation is –
3I 2 (s) + 6 OH – 
5I – + 
+ 3H 2 O( λ )
Δ G° = 5×(–50) + (– 123.5) + 3×(–233)×
–0 – 6× (–150) = – 172.5 KJ/mol
Now
Δ G° = – RT λ nk
⇒ – 172.5 = 
× 300 × 2.3 × 10 –3 log K
⇒ log K = 30
⇒ 10 30 = 
⇒ [OH – ] = 10 –6 ⇒ pOH = 6
⇒ pH = 8
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