Calculate the pH at which the following conversion (reacton ) 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 and `[IO_(3)^(-)] = 0.10M` Given that `Delta G_(f)^(0) (I^(-)aq) = -50` kJ/mole, `Delta G_(f)^(0) (IO_(3)^(-), aq) = -123.5 kJ//"mole", Delta G_(f)^(0) (H_(2)O, l) = - 233` kJ/mole, `DeltaG_(f)^(0) (OH^(-), aq) = - 150` kJ/mole, Ideal gas constant `= R = (25)/(3)J "mole"^(-1) K^(-1)`, log e = 2.3,

Answer

8

Explanation

The balanced reaction is `3I_(2) + 6OH^(-) hArr 5I^(-) + IO_(3)^(-) + 3H_(2)O`
`Delta G^(@)` of reaction `= 5 Delta G_(f)^(@) I^(-) + Delta G_(f)^(@) IO_(3)^(-) + 3 Delta G_(f)^(@) H_(2)O- 3 Delta G_(f)^(@) I_(2) - 6 Delta G_(f)^(@) OH^(-)`
`= 5(-50) + (-123.5) + 3(-233) - 3(0) - 6(-150)`
`= - 250 - 123.5 - 699 + 900 = - 172.5 kJ`
`Delta G^(@) = - 2.303RT log kp`
`kp = 1.452 xx 10^(30)`
So, `1.452 xx 10^(30) = ((0.1)^(5) xx 0.1)/([OH^(-)]^(6))`
`[OH^(-)]^(6) = 6.886 xx 10^(-37)`
`[OH^(-)] = 9.39723 xx 10^(-7)`