Chemistry - pH of solution

  • Thread starter tua28494
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  • #1
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This isnt 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 -> NH4SUB]+/SUP] + OH-/SUP]
the Kb/SUB] =1.5 x 10-3SUP]
 

Answers and Replies

  • #3
10
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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
 
  • #4
Borek
Mentor
28,808
3,308
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?

--
methods
 

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