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Author Topic: What's the concentration of each species present in a solution of .17M phenol?  (Read 3569 times)

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sjbyrne

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I am looking for H3O+, C6H5O-, OH-, and C6H5OH.  I already found Ka which is 1.3x10^-10 and the pH of the solution which is 5.33.  I also found the concentrations of H3O+ and C6H5O- to be 4.7x10^-6 and the concentration of C6H5OH to be (.17) - (4.7x10^-6) but since x is negligible, I dropped it and put the concentration of phenol to be .17M.  The thing I am confused with is finding the concentration of OH-.  When I write the equation, I have C6H5OH + H2O --> C6H5O- + H3O+ but there is no OH- that I see, but there is obviously OH- somewhere because when I put the answer as 0, that is incorrect.  Can someone help me find the concentration of OH- in this solution?
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AWK

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always [OH-][H3O+]=Kw
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sjbyrne

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So if I am trying to find the concentration of all species present (Na+, C6H5CO2-, C6H5CO2H, H3O+, and OH-) in a solution of 0.042M sodium benzoate (C6H5CO2Na) and I know that Ka for benzoic acid is 6.5*10^-5, then do I just take Kw/Ka=Kb and then set Kb=x^3/0.042 to find the concentrations of Na+, C6H5CO2H, and OH- (which I get to be 1.8*10^-4)?  And then to find both H3O+ and C6H5CO2- do I just take Kw/(1.8*10^-4)?

By the way, I get the reaction equation to be C6H5CO2Na + H2O --> Na+ + C6H5COH + OH-
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AWK

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Quote
Kb=x^3/0.042
???

The concentration of Na+ is equal to the concentration of sodium benzoate
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