InterviewSolution
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InterviewSolution
| 1. |
Calculate `pH` of a. `0.002 N CH_(3) COOH` having `2.3%` dissociation. b. `0.002N NH_(4) OH` having `2.3%` dissociation. |
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Answer» a. `0.002 N CH_(3)COOH`: Acetic acid is weak electrlyte and partially associated. `{:(,CH_(3)COOHhArr,CH_(3)COO^(Theta)+,H^(o+),),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}` `:. [H^(o+)] = C alpha = 2 xx 10^(-3) xx (2.3)/(100) = 4.6 xx 10^(-5) M` `:. pH = -log[H^(oplus)] = - log(4.6 xx 10^(5))` `pH = 4.3372` b. `0.002 N NH_(4)OH: NH_(4)OH` is weak base and partially dissociated. `{:(,NH_(4)OHhArr,NH_(4)^(Theta)+,overset(Theta)OH,),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}` `:. [overset(Theta)OH] = C alpha = 2 xx 10^(-3) xx (23)/(100) = 4.6 xx 10^(-5) M` `because pOH = log [overset(Theta)OH] =- log (4.6 xx 10^(-5))` `:. pOH =- 4.3372 :. pH = 14 - 4.3372` `pH = 9.6628` |
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