1.

The ionization constant of acetic acid is 1.74 times 10^-9. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solutions and its pH.

Answer»

SOLUTION :`CH_3COOH leftrightarrowCH_3COO^(-) +H^+`
`K_a=([CH_3COO^(-)][H^+])/([CH_3COOH])=[H^+]^2/([CH_3COOH])`
(or) `[H^+]=SQRT(K_a(CH_3COOH))=sqrt((1.74 times10^-5)(5times10^-2))=9.33 times10^-4M`
`[CH_3COO^(-)]=[H^+]=9.33 times10^-4M`
`pH=-log(9.33 times10^-4)=4-0.9699=4-0.97=3.03`


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