Asked by Hannah
                What is the pH of 0.144M aqueous solution of ammonium nitrate?
I am not sure what to do. Would I do 10^-0.144^ = 0.71?
            
        I am not sure what to do. Would I do 10^-0.144^ = 0.71?
Answers
                    Answered by
            DrBob222
            
    No. You go through the hydrolysis equation. 
.........NH4^+ + H2O ==> H3O^+ + NH3
initial..0.144............0.......0
change....-x...............x......x
equil...0.144-x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute for NH3 and H3O^+ in the above. Substitute 0.144-x for NH4 and solve for x = (H3O^+), then convert to pH.
    
.........NH4^+ + H2O ==> H3O^+ + NH3
initial..0.144............0.......0
change....-x...............x......x
equil...0.144-x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute for NH3 and H3O^+ in the above. Substitute 0.144-x for NH4 and solve for x = (H3O^+), then convert to pH.
                    Answered by
            Hannah
            
    so the values for both NH3 and H3O are 0? 
    
                    Answered by
            DrBob222
            
    No. They are x. And you're solving for x. And you convert x to pH.
    
                    Answered by
            Hannah
            
    For the Ka of NH4^+^ I did 1.0e-14/1.83-5 = 5.55e-10. What do I do with this value?
So for (NH3)(H3O) / NH4 it would be
x^2/0.144-x
    
So for (NH3)(H3O) / NH4 it would be
x^2/0.144-x
                    Answered by
            DrBob222
            
    Kb = 5.55E-10  so
5.55E-10 = x^2/0.144-x
    
5.55E-10 = x^2/0.144-x
                    Answered by
            Hannah
            
    oh ok I get it, thank you!!
    
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