[Rutherford]

Will a reaction occur between NH₄Cl and NaOH in a solution? I think not because both are soluble in water and the substances that are produced are soluble in water. In my book there is a reaction  .

[AWK]

Does a strong base replaces a weaker one in the salt?

[Rutherford]

The reaction is:
NH₄Cl+NaOH-->NaCl+NH₄OH
Is NH₄OH less soluble than NaOH?

[Arkcon]

The reaction is:
NH₄Cl+NaOH-->NaCl+NH₄OH
Is NH₄OH less soluble than NaOH?

Does either have an odor?  What does "having a odor" mean?  Reactions are not driven forward only by a precipitate, gas evolution also works in that regard.

[Borek]

It is not about solubility, it is about acid-base reaction.

[Rutherford]

As I understood, NH₄OH is soluble in water (in physical way), but it doesn't dissociate so fast as NaOH. Or maybe you suppose that NH₃ is made, but it's opposite to my book answer.

[Borek]

What is the book answer?

[Rutherford]

I will write the whole problem and the book answer:
Calculate the pH of a solution that is made by mixing 200cm³ of NH₄Cl (c=4mol/dm³) and 2dm³ of NaOH (c=0.2mol/dm³). Kb=2*10^-5
Answer:
After the mixing: c(NH₄Cl)=0.36mol/dm³, c(NaOH)=0.18mol/dm³.
NH₄Cl+NaOH-->NaCl+NH₄OH (this confuses me)
0.36-0.18=0.18            0.18   

Now they calculated it as a buffer and got pOH, and then pH.

[AWK]

Half of NH₄Cl reacts with NaOH forming ammonia forming buffer solution. In your case pOH = pK_b (-log)K_b) ), and pH=14-pK_b. You should show this using calculations.

[Borek]

Think about it this way - you start with a solution containing NH₄⁺. When you add strong base, you add OH^-. NH₄⁺ is a weak acid, that reacts with OH^- yielding ammonia:

NH₄⁺ + OH^- -> NH₃ + H₂O

Use stoichiometry to calculate concentrations of both NH₄⁺ and NH₃, this is your buffer.

[Rutherford]

That seems more reasonable to me. Maybe the book answer is simplified. Thanks for the help.

[Borek]

Book answer lists NH₄OH instead of NH₃, that's the main difference.

[Rutherford]

But, I think that the book reaction is not correct. The reaction you wrote seems correct to me:
NaOH+NH₄Cl-->NH₃+NaCl+H2O
NH₃ and NH₄⁺ that left make a buffer. Then the NH3 that is made isn't a gass, right?
And one more question: how would NH₄OH alone react in a solution, NH₄⁺ is a weak acid and OH^- is a base?

[Borek]

Product of the reaction is NH₃(aq). It is sometimes written as NH₄OH (there is plenty of water around), but from what I remember existence of this compound is rather dubious.

Ammonia is pretty well soluble in water, but yes, such solutions - if left open - will slowly loose the ammonia.

As presence of NH₄OH is dubious, there is not much sense in discussing its reactions. Note that it is possible we in fact deal with the multistep equilibrium

NH₃ + H₂O <-> NH₄OH <-> NH₄⁺ + OH^-

[Rutherford]

Thanks for the good explanation. I will just stick to the reaction you posted before.