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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on March 24, 2012, 12:37:13 PM

Title: NH4Cl+NaOH
Post by: Rutherford on March 24, 2012, 12:37:13 PM
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  ??? .
Title: Re: NH4Cl+NaOH
Post by: AWK on March 24, 2012, 01:34:17 PM
Does a strong base replaces a weaker one in the salt?
Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 24, 2012, 02:04:54 PM
The reaction is:
NH4Cl+NaOH-->NaCl+NH4OH
Is NH4OH less soluble than NaOH?
Title: Re: NH4Cl+NaOH
Post by: Arkcon on March 24, 2012, 02:24:51 PM
Quote from: Raderford on March 24, 2012, 02:04:54 PM
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.
Title: Re: NH4Cl+NaOH
Post by: Borek on March 24, 2012, 02:26:43 PM
It is not about solubility, it is about acid-base reaction.
Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 24, 2012, 02:50:45 PM
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.
Title: Re: NH4Cl+NaOH
Post by: Borek on March 24, 2012, 05:57:34 PM
What is the book answer?
Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 25, 2012, 03:28:40 AM
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.
Title: Re: NH4Cl+NaOH
Post by: AWK on March 25, 2012, 03:47:20 AM
Half of NH4Cl reacts with NaOH forming ammonia forming buffer solution. In your case pOH = pKb (-log)Kb) ), and pH=14-pKb. You should show this using calculations.
Title: Re: NH4Cl+NaOH
Post by: Borek on March 25, 2012, 04:42:43 AM
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.
Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 25, 2012, 05:10:51 AM
That seems more reasonable to me. Maybe the book answer is simplified. Thanks for the help.
Title: Re: NH4Cl+NaOH
Post by: Borek on March 25, 2012, 06:16:16 AM
Book answer lists NH4OH instead of NH3, that's the main difference.
Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 25, 2012, 07:22:48 AM
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?
Title: Re: NH4Cl+NaOH
Post by: Borek on March 25, 2012, 07:34:29 AM
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-

Title: Re: NH4Cl+NaOH
Post by: Rutherford on March 25, 2012, 07:50:00 AM
Thanks for the good explanation. I will just stick to the reaction you posted before.