Coupled Equilibria – Multiple Equilibria Example

Multiple Equilibria
Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion $Ag(S_2O_3)_2^{3-}$ $(K_f=4.7×10^{13})$.

What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $Ag(S_2O_3)_2^{3-}$?

Two equilibria are involved when AgBr dissolves in a solution containing the $Ag(S_2O_3)_2^{3-}$ ion:


$$AgBr(s)⇌Ag^+(aq)+Br^-(aq)\qquad K_{sp}=5.0×10^{-13}$$


$$Ag^+(aq)+2S_2O_3^{2-}(aq)⇌Ag(S_2O_3)_2^{3-}(aq)\qquad K_f=4.7×10^{13}$$


First, calculate the concentration of bromide that will result when the 1.00 g of AgBr is completely dissolved via the cited complexation reaction:

$$\frac{1.00\;g\;AgBr}{(187.77\;g/mol)}\times\frac{(1\;mol\;Br^-)}{(1\;mol\;AgBr)}=0.00532\;mol\;Br^-$$ $$0.00532\;mol\;Br^-/1.00\;L=0.00532\;M\;Br^-$$

Next, use this bromide molarity and the solubility product for silver bromide to calculate the silver ion molarity in the solution:


Based on the stoichiometry of the complex ion formation, the concentration of complex ion produced is


Use the silver ion and complex ion concentrations and the formation constant for the complex ion to compute the concentration of thiosulfate ion.

$$[S_2O_3^{2-}]^2=[Ag(S_2O_3)_2^{3-}]/[Ag^+]\quad K_f=0.00532/(9.6×10^{-11})(4.7×10^{13})=1.18×10^{-6}$$ $$[S_2O_3^{2-}]=1.08×10^{-3}\;M$$

Finally, use this molar concentration to derive the required mass of sodium thiosulfate:


Thus, 1.00 L of a solution prepared from 1.7 g Na2S2O3 dissolves 1.0 g of AgBr.

Check Your LearningAgCl(s), silver chloride, has a very low solubility:

$$AgCl(s)⇌Ag^+(aq)+Cl^-(aq),\qquad K_{sp}=1.6×10^{-10}. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:

$Ag^+(aq)+2NH_3(aq)⇌Ag(NH_3)_2^+(aq),\quad K_f=1.7×10^7$. What mass of NH3 is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $Ag(NH_3)_2^+$?



1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of AgCl.