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 $Ag(S_2O_3)_2^{3-}$?

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

dissolution:

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

complexation:

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

 

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:

$$[Ag^+]=K_{sp}/[Br^-]=5.0×10^{-13}/0.00532=9.4×10^{-11}\;M$$

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

$$0.00532-9.64×10^{-11}=0.00532\;M$$

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:

$$(1.08×10^{-3}\;mol\;S_2O_3^{2-}/L)(1\;mol\;NaS_2O_3/1\;mol\;S_2O_3^{2-})(158.1\;g\;Na_2S_2O_3/mol)$$

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 $Ag(NH_3)_2^+$?

 

Answer:

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