[antonioo1995]

 Hi everybody 

In order to receive more answers, I will show my attempts to work out the exercise this time 

Once again, sorry for my bad English

So,
500 ml of a 0.2 M solution of NH3 are added to 500 ml of a 0.2 M solution of HCl.
Calculate the pH of the resulting solution . (Kb NH3 = $$1.75 x 10^{-5} mol / L$$ )

first of all:
HCl + NH3  NH4 + Cl

NH4 gives acidic hydrolysis, then:
NH4 + H2O  NH3 + H3O

let's calculate HCl's moles:
$$(0.2 mol/L ) x (0.500L) = 0.1 mol$$, the same as NH3

now I want to know the Ka of the conjugate acid of NH3, so NH4:
$$Ka=Kw/Kb = 5.71x10^{-10}$$

So, since the amount of H+ produced by the reaction is exactly the amount of H3O+ produced by the hydrolysis of NH4 ( or at least I think  ), I need the molarity of NH4.
Let's call [NH4] : y

_____________________
NH4     NH3   H3O+
y            0        0
-x           x        x
y-x         x        x
____________________


$$(x^2)/(y-x) = 5.71x10^{-10}$$

Now x is the molarity of H3O and then I can easily calculate the pH.

My own question is: how can I know the molarity of NH4? 

thanks guys

[Borek]

How many moles of NH₄⁺ were produced, and what was the final volume of the solution?

[antonioo1995]

You need more data or it's just a question for me? 
 
The final volume of the solution is 1L.
As for the moles of NH4, I don't know...it's what I've asked for ( M=mol/L) 

[Borek]

or it's just a question for me? 

Question for you.
 

The final volume of the solution is 1L.

OK

As for the moles of NH4, I don't know...it's what I've asked for ( M=mol/L) 

Come on, this is a simple stoichiometry. How many moles of NH₃ reacted? You already know all of it reacted, don't you?

Forget about the equilibrium for now.

[antonioo1995]

0.1 moles of NH3 reacted.
But why should 0.1 mol of NH4 react too? Only HCl is completely ionized, NH3 is a weak base so I don't understand why moles of NH3 = moles NH4

I thought stechiometry was  useful only when strong acids or bases (solo) are present. On the other hand when there is a weak acid or base I thought I had to complete that pattern used in my question for the calculus of NH4's moles 

[Borek]

0.1 moles of NH3 reacted.
But why should 0.1 mol of NH4 react too? Only HCl is completely ionized, NH3 is a weak base so I don't understand why moles of NH3 = moles NH4

How is the final solution different from one prepared dissolving just NH₄Cl in appropriate amount of water?

I never said 0.1 mol of NH₄⁺ reacted. What I am telling you is that you have a solution of NH₄⁺ of a known initial concentration, and NH₄⁺ is a weak acid.

[antonioo1995]

You said " it's a simple stoichiometry".
And looking at the stoichiometry of the reaction what I notice is that moles of NH3 ( 0.1 moles ) = moles NH4

But, let me think...
Ok, now what you say is that the reaction between NH3 and HCl gives a salt and water as products following this reaction:

HCl + NH3  NH4Cl + H2O

And this thing rises me an other doubt...why should a strong acid and a weak base give a salt + water? I thought that happened only when strong base and strong acid are concerned and that this fact was called neutralization 

Let's switch to the other question:
if I dissolve NH4Cl in water, I will receive an acidic solution cause Cl is the conjugate of a strong acid...but I have still no idea on how to calculate the amount of NH3...

[mjc123]

first of all:
HCl + NH3  NH4 + Cl

NH4 gives acidic hydrolysis, then:
NH4 + H2O  NH3 + H3O

You are over-complicating things. You don't need two equations. HCl is a strong acid, so is fully ionised. Cl^- is a spectator ion, it doesn't participate in the reaction. So you can just write the reaction as
H₃O⁺ + NH₃  NH₄⁺ + H₂O
Now you just have one equation to deal with. Does that make it simpler?

[antonioo1995]

Why is H3O in the reagents? I mean, this reaction gives an acidic solution so why H3O isn't in the products?

[antonioo1995]

oh wait, maybe I've understood:

H + Cl + NH3 + H20 NH4 + Cl +H20

Cl are spectators so

 NH3  + H3O NH4 + H20

0.1mol     0                  0
0.1mol   -0.1mol       0.1mol
0             0               0.1mol

[NH4] = 0.1M

NH4 + H20  NH3 +H30
0.1M                          o       0
-x                              x        x
0.1M-x                       x         x

$$x=[H3O]$$
$$Ka = 5.7 x 10^{-10}$$
$$x = \sqrt[2]{Kax0.1}$$
$$x= 7.5 x 10^{-6}$$

$$pH = 5.12$$

right? 

[Borek]

pH of 5.12 looks OK.

You are still overcomplicating things, you don't ICE table for a first step (stoichiometry), then the way you calculate x is based on ignoring the ICE table content - so why do you construct it?

To get multiplication sign in LaTeX use \times not x.

[tex]K_a = 5.7\times 10^{-10}[/tex]

[antonioo1995]

Yeah the second table is redundant, I know 

[Borek]

Yeah the second table is redundant, I know 

No, BOTH are redundant if you directly use the formula you used.

Have you checked if using it is OK?

[antonioo1995]

Well, probably you are right, but I see the first one as a way to better organize my ideas