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The dissolution of ammonia gas in water ...

The dissolution of ammonia gas in water does not obey Henry. s law. On dissolving , a major portion of ammonia molecules unite with `H_2O ` to form `NH_4 OH` molecules . A portion of the latter again dissociates into ` NH_4^(+) and OH ^(-) ` ions. In solution therefore , we have `NH_3` molecules, `NH_4OH` molecules and `NH_4^(+) ` ions and the following equilibrium exist :
` NH_(3(g) ) ` (pressure P and concentration c) `hArr NH_(3(i)) +H_2O hArr NH_4 OH hArr NH_4^(+) +OH^(-) ` Let `c_1 ` mol//L of `NH_3` pass in liquid state which on dissolution in water forms `c_2 `mol/ L of `NH_4 OH. ` The solution contains `c_3` mol/L of `NH_4^(+) ` ions.
The dissociation constant of `NH_4 OH ` can be given as :

A

` K_b =((c_3)^(2))/((c_2-c_3))`

B

` K_b =((c3)^(2))/(c_2)`

C

` K_b =(c_2)/( (c_2-c_3))`

D

` K_b= ( c_3)/((c_1-c_2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`K_b =([NH_4^(+)][OH^(-)])/( [NH_4OH]) = ( (C_3)(C_3))/( C_2-C_3)`
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