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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` |
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Answer» Correct Answer - pH=8 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` |
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