Calculate the pH at which the following reaction will be at equilibrium in basic medium :
`A_(2)(s) hArr A^(-)(aq) +AO_(3)^(-) (aq)`
When the equilibrium concentration at 300 K are:
`[A^(-)] = 0.1M, [AO_(3)^(-)] = 0.1M`
Given: `DeltaG_(f)^(@)` of `OH^(-)(aq) =- 150 kJ//"mole"`,
`DeltaG_(f)^(@) "of" H_(2)O(l) =- 233 kJ//"mole"`
`DeltaG_(f)^(@) "of" A^(-)(aq) =- 50 kJ//"mole"`,
`DeltaG_(f)^(@) "of" AO_(3)^(-) (aq) =- 123.5 kJ//"mole"`,
`R = (25)/(3) J "mole"^(-1) K^(-1), log_(e) 10 = 2.3`

Answer

pH=8

Explanation

Balanced equation will be
`3I_2(s)+6OH^(-)hArr5I^(-)(aq)+IO_3^(-) (aq)+3H_2O(l)`
`DeltaG^@=5xx(-50)+(-123.5)+3xx(-233)-0-6xx(-150)`
`DeltaG^@= -172.5 kJ"mole"^(-1) =-25/3xx300xx2.3xx10^(-3)` log k
log k =30
`10^30=(10^(-5)xx10^(-1))/([OH^(-)]^6)`
`:. [OH^(-)]=10^(-6)`
`:. pH=14-6=8`