How Do You Calculate the pH of a Mixed NH4OH and NH4Cl Solution?

[tua28494]

This isn't really homework, Its just a problem I'm trying to learn to solve. I'm not taking a chemistry course so I'm learning on my own.


Calculate the pH of a solution that is 0.65 M NH₄OH and 0.35 M NH₄Cl.

Well we can calulate the pOH = -log(OH^-) and 14-pOH = pH


I guess I need the equilibrium concentration of OH^-.

here is the chemical formula. I believe is necessary

NH₄OH -> NH_{4SUB]^{+/SUP] + OH-/SUP]
the Kb/SUB] =1.5 x 10-3SUP]}}

 

[Borek]

This is a buffer solution, so you should use Henderson-Hasselbalch equation.

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methods

 

[tua28494]

Without using the Henderson–Hasselbalch equation, Can I do this?
also different Kb value.

The NN4CL will completely dissociate adding 0.35 NH4

NH4OH -> NH4+ + OH-
0.65-x. 0.35+x. . x

H2O + NH3 -> NH4+ + OH-

Kb =1.8 x 10^-5 = [NH4+][OH-]/[NH3]= (0.35+x)(x)/ 0.65-x

x = [OH-]=0.000033 M

pOH = 4.5
pH = 14 - pOH = 14 - 4.5 =9.5

 

[Borek]

This is correct by accident.

You look at the dissociation of NH₄OH, but you ignore acidic dissociation of NH₄⁺.

NH₄⁺ <-> NH₃ + H⁺

Why?

My bet is that if you will try to do calculations using your approach and based on this reaction, you will get almost the same result, again accidentally.

For very small x (compared with 0.35 & 0.65), is (0.35+x)/(0.65-x) substantially different from 0.35/0.65?

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methods