A solution of acetic acid is 1% ionized. determine the concentration of the H3O positive of the solution. Ionization constant is 1.75*10 to the negative fifth power?
The chemical equation of this reaction can be written as:
`CH_3COOH + H_2O -> CH_3COO^(-) + H_3O^+`
and the ionization constant is given as:
`K_a = ([H_3O^+][CH_3COO^-])/([CH_3COOH])`
% ionization is given as:
`% ionization = ([H_3O^+])/([CH_3COOH]) xx 100 = 1`
`[H_3O^+] = 0.01 [CH_3COOH]`
Using stoichiometry, concentrations of the products is in 1:1 ratio and using the % ionization data, concentration of Hydronium ion is available in terms of acid concentration. Let us assume concentration of hydronium ion is x.
Using these relations in the equation for Ka, we get
`Ka = x xx x/([CH_3COOH]) = 0.01 xx 0.01 [CH_3COOH]`
or, [CH3COOH] = 10^4 Ka = 10^4 x (1.75 x 10^-5) = 0.175 M
and [H3O+] = 0.01 [CH3COOH] = 0.00175 M = `1.75 xx 10^(-3) M`
Hence at 1% ionization level, hydronium ion and acetate ion is 1.75 x10^-3 M, and the initial given concentration of acetic acid is 0.175 M.
Hope this helps.