Correct Answer - `[Ac^(-)] = 0.00093, pH = 3.03`
Method 1
`{:(1,CH_(3)COOHharr,CH_(3)COO^(-)+H^(+),K_(a)=1.74xx 10^(-5)),(2,H_(2)O+H_(2)Oharr,H_(3)O^(+)+OH^(-),K_(w) = 1.0 xx 10^(-14)):}`
Since `Ka gtgt K_(w)`, :
`{:(,CH_(3)COOH,+H_(2)Oharr,CH_(3)COO^(-),+H_(3)O^(+)),(C_(t)=,0.05,,0,0),(,0.05-.05alpha,,0.05alpha,0.05alpha):}`
`K_(a) = ((.05 alpha) (.05alpha))/((.05-0.05alpha))`
` = ((.05alpha)(0.05alpha))/(.05(1-alpha))`
`(.05alpha^(2))/(1-alpha)`
`1.74 xx 10^(-5) = (0.05alpha^(2))/(1-alpha)`
`1.74 xx 10^(-5) - 1.74 xx 10^(-5) alpha = 0.05alpha^(2)`
`0.05alpha^(2) + 1.74 xx 10^(-5) alpha - 1.74 xx 10^(-5)`
`D = b^(2) - 4ac`
`= (1.74 xx 10^(-5))^(2) - 4(.05) (1.74 xx 10^(-5))`
`= 3.02 xx 10^(-25) + .348 xx 10^(-5)`
`alpha = sqrt((K_(a))/(c))`
`alpha = sqrt((1.74 xx 10^(-5))/(.05))`
`=sqrt(3.48 xx 10^(-6))`
`= CH_(3)COOH harr CH_(3)COO^(-) + H^(+)`
`alphaunderline(1.86xx10^(-3))`
`[CH_(3)COO^(-)] = 0.05 xx 1.86 xx 10^(-3)`
`= (0.93 xx 10^(-3))/(1000)`
`= .00093`
Method 2
Degree of dissociation,
`alpha = sqrt((K_(a))/(c))`
`c = 0.05 M`
`K_(a) = 1.74 xx 10^(-5)`
Them, `alpha = sqrt((1.74 xx 10^(-5))/(.05))`
`alpha = sqrt(34.8 xx 10^(-5))`
`alpha = sqrt(3.48) xx 10^(-4)`
`alpha = 1.8610^(-2)`
`CH_(3)COOH harr CH_(3)COO^(-) + H^(+)`
Thus, concentration of `CH_(3)COO^(-) = c.alpha`
`= .05 xx 1.86 xx 10^(-2)`
`= .093 xx 10^(-2)`
`= .00093 M`
Since `[oAc^(-)] = [H^(+)]`,
`[H^(+)] = .00093 = .093 xx 10^(-2)`
`pH = - "log" [H^(+)]`
`= - "log" (.093 xx 10^(-2))`
`:. pH = 3.03`
Hence the concentration of acetate ion in the solution is `0.00093` M and its Ph is `3.03`.
