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Question

The ionization constant of acetic acid is 1.74×105. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH. 



    Solution

    The correct option is

    Method 1
    1) CH3COOHCH3COO+H+   Ka=1.74×105
    2) H2O+H2OH3O++OH     Kw=1.0×1014
    Since Ka>>K2:
    CH3COOH+H2OCHCOO+H3O+c1=0.05000.05.05α0.05α0.05α
    Ka=(.05α)(.05α)(.05α0.05α)=(.05α)(0.05α).05(1α)=.05α21α1.74×105=0.05α21α1.74×1051.74×105α=0.05α20.05α2+1.74×105α1.74×105D=b24ac=(1.74×105)24(.05)(1.74×105)=3.02×1025+.348×105α=Kacα=1.74×105.05=34.8×105×1010=3.48×106=CH3COOHCH3COO+H+α1.86×103[CH3COO]=0.05×1.86×103=0.93×1031000=.000093

    Method 2
    Degree of dissociation, 
    α=KacC=0.05MKa=1.74×105
    Then, α=1.74×105.05 α=34.8×105α=3.48×104α=1.86102CH3COOHCH3COOHCH3COO+H+
    Thus, concentration of CH3COO=C.a
    =0.5×1.86×102=.093×102=.00093M
    Since [oAc]=[H+][H+]=.00093=.093×102pH=log[H+]=log(.093×102)pH=3.03
    Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03. 


    Chemistry
    NCERT Textbook
    Standard XI

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