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 $Na_3[Ag(S_2O_3)_2]$?

Md Ariful Islam

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Md Ariful Islam

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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 $"silver bromide"$, and moles of $"sodium thiosulfate"$, and $1:1$ stoichiometric equivalence between moles of $"silver bromide"$, and moles of $"silver(I) thiosulfate anion"$.

$"Moles of silver bromide"=(0.22*g)/(187.77*g*mol^-1)=1.17xx10^-3*mol$.

Clearly there is a $1.17xx10^-3*mol$ quantity of $Na_3[Ag(S_2O_3)_2]$.

And this represents a mass of .......................... $1.17xx10^-3*molxx401.11*g*mol^-1~=0.50*g$

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