Equilibrium Question 913
Question: The number of moles of $ NH_3 $ that must be added to 1 L of 0.1 M $ AgNO_3 $ to reduce $ A{g^{+}} $ concentration to $ 2\times {10^{-7}}M $ are Given, $ K_{dis}{{[Ag{{(NH_3)}_2}]}^{+}}=6.8\times {10^{-8}} $
Options:
A) 0.184 M
B) 0.384 M
C) 0.293 M
D) 0.0539 M
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{[Ag,{{(NH_3)}_2}]}^{+}}=0.1,M $
$ [A{g^{+}}]=2\times {10^{-7}} $
$ {{[Ag{{(NH_3)}_2}]}^{+}}A{g^{+}}+2NH_3 $
$ K_{dis}=\frac{[A{g^{+}}]{{[NH_3]}^{2}}}{{{[Ag{{(NH_3)}_2}]}^{+}}} $
$ 6.8\times {10^{-8}}=\frac{2\times {10^{-7}}\times {{[NH_3]}^{2}}}{0.1} $
$ [NH_3]=0.184,M $
$ {{[NH_3]} _{total}}={{[ NH_3 ]} _{free}}+{{[NH_3]} _{complexed}} $
$ =0.184+2\times 0.1=0.384,M $
