Question

Consider a solution containing 0.181 M lead ions and 0.365

Answer 1

Answer:

$[H_3O^+]=6.8*10^{-22}$

Explanation:

For a solution with $[Pb^{+2}]=1*10^{-6}$ :

$Kps=[Pb^{+2}]*[S^{-2}]=3.4*10^{-28}$

$1*10^{-6}*[S^{-2}]=3.4*10^{-28}$

$[S^{-2}]=3.4*10^{-22}$

For the iron ions:

$Kps=[0.365]*[S^{-2}]=3.7*10^{-19}$

$[S^{-2}]=1.01*10^{-18}$

Given that the $[S^{-2}]$ concentration required to start precipitating iron is higher than the final concentration for the Pb precipitation (when the desired concentration is reached), the iron will not precipitate.

The concentration of $H_3O^+$

$H_2S + 2 H_2O \longrightarrow 2 H_3O^+ + S^{-2}$

$[H_3O^+]=2*[S^{-2}]=6.8*10^{-22}$