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)
A. A
XS(s) X^ +2 (aq.) + S^ 2- (aq.) For precipitation of XS(s) [X^ +2 ][S^ 2- ] K_ sp (XS) [S^ 2- ] 1 10^ -22 0.01 = 10^ -20 YS(s) Y^ +2 (aq) + S^ 2- (aq) For precipitation of YS(s) [Y^ +2 ][S^ 2- ] K_ sp (YS) [S^ 2- ] 4 10^ -16 10^ -2 = 4 10^ -14 Now, H_2S(aq) 2H^+(aq) + S^ 2- (aq) [S^ 2- ][H^+]^2 H_2S = K_ a_1 K_ a_2 = 1 10^ -21 [S^ 2- ] = 1 10^ -21 [H_2S] [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.
Related: Chemistry — Ionic Equilibrium · All PYQ Banks