1. Tardigrade
  2. Question
  3. Chemistry
  4. The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia, molecules unite with H 2 O to form NH 4 OH molecules. NH 4 OH again dissociate into NH 4+and OH -ions. In solution therefore, we have NH 3 molecules, NH 4 OH molecules and NH 4+ions and the following equilibrium exist: NH 3( g ) (pressure p and concentration C ) initially leftharpoons NH 3(l)+ H 2 O leftharpoons NH 4 OH leftharpoons NH 4++ OH - Let C1 mol / L of NH 3 pass in solution state a part of which on dissolution in water forms C2 mol / L of NH 4 OH. The solution contains C3 mol / L of NH 4+ions Total concentration of ammonia, which can be determined by volumetric analysis is equal to:

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Q. The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia, molecules unite with $H _2 O$ to form $NH _4 OH$ molecules. $NH _4 OH$ again dissociate into $NH _4^{+}$and $OH ^{-}$ions. In solution therefore, we have $NH _3$ molecules, $NH _4 OH$ molecules and $NH _4^{+}$ions and the following equilibrium exist:
$NH _3( g )$ (pressure $p$ and concentration $C$ ) initially $\rightleftharpoons NH _3(l)+ H _2 O \rightleftharpoons NH _4 OH \rightleftharpoons NH _4^{+}+ OH ^{-}$
Let $C_1 \, mol / L$ of $NH _3$ pass in solution state a part of 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
Total concentration of ammonia, which can be determined by volumetric analysis is equal to:

Equilibrium

Solution:

The intermediate solution of acid will react with all the $NH_3 $ present in solution