1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chemistry - pH of solution

  1. Apr 12, 2010 #1
    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]
     
  2. jcsd
  3. Apr 12, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

  4. Apr 12, 2010 #3
    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
     
  5. Apr 12, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Chemistry - pH of solution
Loading...