Question

When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. the solution is found to have a hydroxide ion concentration of 1.34 × 103. the dissociation constant of ammonia is?

Answer 1

Reaction involved-
NH3  +  H2O  → NH4+  +  OH-

Given-
NH4+ = OH-  = 1.34×10-3 M
NH3 reacted = (0.1 -1.34 ×10-3) M

Condition at equilibrium-
= (0.1 -1.34 ×10-3) M 
= 0.1 - 0.00134 M
= 0.09866 M 
 
Kb 
= [NH4+ ][OH- ]/[NH3 ]
= (1.34×10-3 )× 1.34×10-3)/0.09866
=1.8 × 10-5

Answer 2

"it seems this is what you are looking for"

When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. the solution is found to have a hydroxide ion concentration of 1.34 × 10⁻³. the dissociation constant of ammonia is?

Answer:

The dissociation constant of the ammonia in the solution will be equal to 1.8×10⁻⁵.

Explanation:

Given that the concentration of  OH⁻ ion,  $C\alpha = 1.34*10^{-3}$

where $C=0.1molL^{-1}$

Then   $\alpha =\frac{1.34*10^{-3}}{0.1} =1.34*10^{-2}$

The chemical equation:

                                NH₃   +   H₂O       ⇆       NH₄⁺     +     OH⁻

At equilibrium       C(1-α)      C(1-α)                   Cα               Cα

The dissociation constant , $K_{d}=C\alpha ^{2}$

             $K_{d}=(0.1)(1.34*10^{-2})^{2}\\ K_{d}=1.8*10^{-5}$

Therefore the dissociation constant of NH₃ will be equal to 1.8×10⁻⁵.