The ionization constant of acetic acid is `1.74 xx 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.

Explanation

`K_(a)=1.74xx10^(-5), pK_(a)=-log (1.74xx10^(-5))`
`=-log 1.74+5`
`=-0.2405+5=4.7595 ~~ 4.76`
`pH_(W_(A))=1/2(pK_(a)-logC)`
`=1/2[4.7595-log (5xx10^(-2))]`
`=1/2(4.76-0.7+2)=3.03`
`pH_(W_(A))=3.03`
ii. `alpha=sqrt(K_(a)/C)=((1.74xx10^(-5))/0.05)^(1//2)=1.865xx10^(-2)M`
iii. `[CH_(3)COO^(Θ)]=1.865xx10^(-2)xx0.05`
`0.09325xx10^(-2)`
`=0.00093 M`