In a qualitative analysis Bi ³⁺ is detected by appearance of precipitate of BiO(OH)(s) . Calculate pH when the following equilibrium exists at 298 K : BiO(OH)(s) BiO ⁺ (aq) + OH ⁻ (aq) , K = 4 10⁻¹⁰ (Given : 2 = 0.3010 )

Step-by-step solution

The equilibrium reaction is given by: BiO(OH)(s) BiO ⁺ (aq) + OH ⁻ (aq) The equilibrium constant expression is: K = [ BiO ⁺][ OH ⁻] Let the solubility of BiO(OH) be s . Then [ BiO ⁺] = s and [ OH ⁻] = s . Substituting the values into the equilibrium expression: s^2 = 4 10⁻¹⁰ s = 2 10⁻⁵ M Therefore, the concentration of hydroxide ions is: [ OH ⁻] = 2 10⁻⁵ M Calculating the pOH: pOH = - [ OH ⁻] = - (2 10⁻⁵) = 5 - 2 Given 2 = 0.3010 : pOH = 5 - 0.3010 = 4.699 The pH of the solution at 298 K is: pH = 14 - pOH = 14 - 4.