Consider two Group IV metal ions X ²⁺ and Y ²⁺ . A solution containing 0.01 M X ²⁺ and 0.01 M Y ²⁺ is saturated with H ₂ ~S . The pH at which the metal sulphide YS will form as a precipitate is _ _ _ _ . (Nearest integer) (Given: K _ sp ( XS )=1 10⁻²² at 25^ C , K _ sp ( YS )=4 10⁻¹⁶ at 25^ C , [ H ₂ ~S ]=0.1 M in solution, K _ a 1 K _ a 2 ( H ₂ ~S )=1.0 10⁻²¹, 2=0.30 , 3=0.48, 5=0.70 )

Step-by-step solution

XS(s) X⁺²(aq.) + S²⁻(aq.) For precipitation of XS(s) [X⁺²][S²⁻] K_ sp (XS) [S²⁻] 1 10⁻²² 0.01 = 10⁻²⁰ YS(s) Y⁺²(aq) + S²⁻(aq) For precipitation of YS(s) [Y⁺²][S²⁻] K_ sp (YS) [S²⁻] 4 10⁻¹⁶ 10⁻² = 4 10⁻¹⁴ Now, H₂S(aq) 2H^+(aq) + S²⁻(aq) [S²⁻][H^+]^2 H₂S = K_ a₁ K_ a₂ = 1 10⁻²¹ [S²⁻] = 1 10⁻²¹ [H₂S] [H^+]^2 4 10⁻¹⁴ [H^+]^2 1 4 10⁻⁷ 10⁻¹ [H^+] 1 2 10⁻⁴ pH 4.3 Nearest integer pH = 4.