JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Ionic Equilibrium

The number of moles of \[N{{H}_{3}}\] that must be added to 1 L of 0.1 M \[AgN{{O}_{3}}\] to reduce \[A{{g}^{+}}\] concentration to \[2\times {{10}^{-7}}M\] are Given, \[{{K}_{dis}}{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}=6.8\times {{10}^{-8}}\]

A) 0.184 M

B) 0.384 M

C) 0.293 M

D) 0.0539 M

Correct Answer: B

Solution :

[b] \[{{[Ag\,{{(N{{H}_{3}})}_{2}}]}^{+}}=0.1\,M\]

\[[A{{g}^{+}}]=2\times {{10}^{-7}}\]

\[{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}A{{g}^{+}}+2N{{H}_{3}}\]

\[{{K}_{dis}}=\frac{[A{{g}^{+}}]{{[N{{H}_{3}}]}^{2}}}{{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}}\]

\[6.8\times {{10}^{-8}}=\frac{2\times {{10}^{-7}}\times {{[N{{H}_{3}}]}^{2}}}{0.1}\]

\[[N{{H}_{3}}]=0.184\,M\]

\[{{[N{{H}_{3}}]}_{total}}={{\left[ N{{H}_{3}} \right]}_{free}}+{{[N{{H}_{3}}]}_{complexed}}\]

\[=0.184+2\times 0.1=0.384\,M\]