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Calculate the pH of a 5.80×10-1 M aqueous solution of sodium formate (NaCO2H). 

(For formic acid, HCO2H, Ka = 1.80×10-4).

Feb 15th, 2015

HCO2 -     +     H2O     =>     HCO2H    +    OH-   ;   Kb = Kw/Ka = (10^-14)/(1.80*10^-4) = 5.56 * 10^-11 

 0.580                                         0                  0

     -x                                          +x                 +x

(0.580-x)                                   (x)                 (x)

So Kb = (5.56*10^-11) = (x^2)/(0.580-x)         => assume 0.580 - x = 0.580 

then x = sqrt((5.56*10^-11)*0.580) = 5.68*10^-6 M = [OH-]

pOH = -log(5.68*10^-6) = 5.3 

and pH + pOH = 14    =>    pH = 14 - pOH = 14 -5.3 = 8.7


Feb 15th, 2015

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