Answer to Question #143974 in General Chemistry for esther

Question

Hello my question is:
Solid ammonium bromide is slowly added to 50.0 mL of a 0.0407 M silver nitrate solution. The concentration of bromide ion required to just initiate precipitation is?

Answer 1

The chemical equation for the reaction is;

$NH_4Br_{(aq)}+AgNO_{3{(aq)}}\to NH_4NO_{3{(aq)}}+AgBr_{(s)}$

The ionic equation for the reaction is;$NH_4^++Br^-+Ag^++NO_3^-\to NH_4^++NO_3^-+AgBr$

$\therefore$ Net ionic equation is;

$Br^-_{(aq)}+Ag^+_{(aq)}\to AgBr_{(s)}$

$K_{sp}of AgNO_3=1.8\times 10^{-18}$

$K_{sp} of AgBr=7.7\times 10^{-13}$

$K_{sp}AgNO_3=[Ag^+][NO_3^-]$

$1.8\times 10^{-18}=[Ag^+]\times 0.0407M$

$[Ag^+]$ $=4.4226\times 10^{-18}M$

$K_{sp}AgBr=[Ag^+][Br^-]$

$7.7\times 10^{-13}=$ $4.4226\times 10^{-18}$ $\times[Br^-]$

$[Br^-]=1.741\times10^4M$

$\therefore$ The concentration of $Br^-=$ $1.741\times10^4M$