Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium ?
`I_2(s)hArr I^(-)(aq)+IO_(3)^(-)(aq)`
When the equilibrium concentrations at 300 K are `[I^(-)]` =0.10 M and `[IO_3^(-)]=0.10 M`
{Given that `DeltaG_f^@(I^(-)aq)`=-50 kJ/mole , `DeltaG_f^(@)(IO_3^(-),aq)`=-123.5 kJ/mole , `DeltaG_f^(@)(H_2O,l)=-233` kJ/mole , `DeltaG_f^(@)(OH^(-),aq)`=-150 kJ/mole , Ideal gas constant =R=`25/3` J `"mole"^(-1)K^(-1),log e =2.3 , `deltaG_f^@` (element, standard state )=0, `DeltaG_r^@`(reaction)`=sumv_pDeltaG_f^@`(products)- `sumv_rDeltaG_f^@`(reactants), where `v_p` and `v_r` are the stochiometric coefficients in the balanced chemiclal equation. }

Answer

8

Explanation

Balanced equation will be
`3I_2(s)+6OH^(-)hArr5I^(-)(aq)+IO_(3)^(-)(aq)+3H_2O(l)`
`DeltaG^(@)=-172.5 "kJmole"^(-1)=-25/3xx300xx2.3xx10^(-3)`log k
log k=30
`10^(30)=(10xx^(-5)xx10^(-1))/([OH^(-)]^(6))`
so `[OH^-]=10^(-6)` and therefore `[H^+]=10^(-8)` so, pH=8