Question

# The ionization constant of acetic acid is 1.74×10−5. 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) CH3COOH↔CH3COO−+H+   Ka=1.74×10−5 2) H2O+H2O↔H3O++OH−     Kw=1.0×10−14 Since Ka>>K2: CH3COOH+H2O↔CHCOO−+H3O+c1=0.05000.05−.05α0.05α0.05α Ka=(.05α)(.05α)(.05α−0.05α)=(.05α)(0.05α).05(1−α)=.05α21−α1.74×10−5=0.05α21−α1.74×10−5−1.74×10−5α=0.05α20.05α2+1.74×10−5α−1.74×10−5D=b2−4ac=(1.74×10−5)2−4(.05)(1.74×10−5)=3.02×10−25+.348×10−5α=√Kacα=√1.74×10−5.05=√34.8×10−5×1010=√3.48×10−6=CH3COOH↔CH3COO−+H+α1.86×10−3[CH3COO−]=0.05×1.86×10−3=0.93×10−31000=.000093 Method 2 Degree of dissociation,  α=√KacC=0.05MKa=1.74×10−5 Then, α=√1.74×10−5.05 α=√34.8×10−5α=√3.48×10−4α=1.8610−2CH3COOH↔CH3COOH↔CH3COO−+H+ Thus, concentration of CH3COO−=C.a =0.5×1.86×10−2=.093×10−2=.00093M Since [oAc−]=[H+][H+]=.00093=.093×10−2pH=−log[H+]=−log(.093×10−2)∴pH=3.03 Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.  ChemistryNCERT TextbookStandard XI

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