A) 0.184 M
B) 0.384 M
C) 0.293 M
D) 0.0539 M
Correct Answer: B
Solution :
[b] \[{{[Ag\,{{(N{{H}_{3}})}_{2}}]}^{+}}=0.1\,M\] |
\[[A{{g}^{+}}]=2\times {{10}^{-7}}\] |
\[{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}A{{g}^{+}}+2N{{H}_{3}}\] |
\[{{K}_{dis}}=\frac{[A{{g}^{+}}]{{[N{{H}_{3}}]}^{2}}}{{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}}\] |
\[6.8\times {{10}^{-8}}=\frac{2\times {{10}^{-7}}\times {{[N{{H}_{3}}]}^{2}}}{0.1}\] |
\[[N{{H}_{3}}]=0.184\,M\] |
\[{{[N{{H}_{3}}]}_{total}}={{\left[ N{{H}_{3}} \right]}_{free}}+{{[N{{H}_{3}}]}_{complexed}}\] |
\[=0.184+2\times 0.1=0.384\,M\] |
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