Asked by Anonymous
                For a solution that is 0.168M NH3 and 0.106M NH4Cl calculate the following.
[OH-]
[NH4+]
[Cl-]
[H30+]
            
        [OH-]
[NH4+]
[Cl-]
[H30+]
Answers
                    Answered by
            DrBob222
            
    pH can be calculated from the Henderson-Hasselbalch equaion. 
pOH from pH + pOH = pKw = 14
OH^- from pOH = -log(OH^-)
H3O^+ from pH = -log(H3O^+)
(Cl^-) = NH4Cl
Post your work if you get stuck.
    
pOH from pH + pOH = pKw = 14
OH^- from pOH = -log(OH^-)
H3O^+ from pH = -log(H3O^+)
(Cl^-) = NH4Cl
Post your work if you get stuck.
                    Answered by
            Anonymous
            
    Using the Henderson-Hasselbalch equation, should i write this:
pH= -log ka + log(0.168/0.106)
PH= -log(5.6*10^-10) + log(0.168/0.106)
PH=9.45
    
pH= -log ka + log(0.168/0.106)
PH= -log(5.6*10^-10) + log(0.168/0.106)
PH=9.45
                    Answered by
            Anonymous
            
    @DrBob222 .......Thanks. It worked.
    
                    Answered by
            DrBob222
            
    Yes, and 9.45 is correct if Ka is correct. I didn't check that.
    
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