01-10-09 ammonia and AgCl in equilibrium

JEE Mains 2015 Discussion Forum › Question Of the Day questions › 01-10-09 ammonia and AgCl in equilibrium

Given : Ag(NH3)₂⁺â†’ Ag⁺ + 2NH₃, K_c = 6.2 Ã— 10^â€“8 and K_sp of AgCl =1.8 Ã— 10^â€“10 at 298 K. If ammonia is added to a water solution containing excess of AgCI(s) only,. Calculate the concentration of the complex in 1.0 M aqueous ammonia..

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6 Answers

3

msp ·2009-10-02 09:44:41

concentration of complex=√K_sp/K_c

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3

msp ·2009-10-03 02:45:30

sir may i post the soln wat i got,so dat u can correct me.

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1357

Manish Shankar ·2009-10-03 03:00:36

Is Kc also in the root in your answer

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msp ·2009-10-03 03:06:51

okie sir,i will do one more time.

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1357

Manish Shankar ·2009-11-21 06:57:31

Answer is âˆš(Ksp/Kc)

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Anirudh Kumar ·2009-12-12 23:10:09

AgCl \leftrightarrow Ag⁺ + Cl ^-

let at time of equlibrium [Ag]= y M
and [Cl]= x = slubility of AgCl

xy= K_sp .................. (1)

now

Ag⁺ + 2 NH₃ \leftrightarrow Ag(NH₃)₂⁺ K=1/K_c

now [Ag(NH₃)₂⁺ ] = x-y M

=> x-y/y= K=1/K_c
neglecting change in conc. of NH₃.

=> x/y= 1/K_c ............ (2)

solving one and two x = \sqrt{\frac{K_{sp}}{K_{c}}}

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Your Answer

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