# How does "silver bromide" react with "sodium thiosulfate"? What mass of sodium thiosulfate is required to react with a 0.22*g mass of "silver bromide" to give the complex ion Na_3[Ag(S_2O_3)_2]?

Aug 16, 2017

We assess the reaction:

$A g B r \left(s\right) \uparrow + 2 N {a}_{2} {S}_{2} {O}_{3} \left(a q\right) \rightarrow N {a}_{3} \left[A g {\left({S}_{2} {O}_{3}\right)}_{2}\right] \left(a q\right) + N a B r \left(a q\right)$

#### Explanation:

This is a complexation reaction, and is still used in black and white photography.

And clearly, given the stoichiometric reaction, there is $1 : 2$ stoichiometric equivalence between moles of $\text{silver bromide}$, and moles of $\text{sodium thiosulfate}$, and $1 : 1$ stoichiometric equivalence between moles of $\text{silver bromide}$, and moles of $\text{silver(I) thiosulfate anion}$.

$\text{Moles of silver bromide} = \frac{0.22 \cdot g}{187.77 \cdot g \cdot m o {l}^{-} 1} = 1.17 \times {10}^{-} 3 \cdot m o l$.

Clearly there is a $1.17 \times {10}^{-} 3 \cdot m o l$ quantity of $N {a}_{3} \left[A g {\left({S}_{2} {O}_{3}\right)}_{2}\right]$.

And this represents a mass of .......................... $1.17 \times {10}^{-} 3 \cdot m o l \times 401.11 \cdot g \cdot m o {l}^{-} 1 \cong 0.50 \cdot g$