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 1
`CH_3COOH ⇌ CH_3COO^- + H^+`
`K_a = ([CH_3COO^-][H^+])/[CH_3COOH] = ([H^+]^2)/[[CH_3COOH]]`
or `[H^+] = sqrt(K_a[CH_3COOH])` = `sqrt((1.74xx10^(-5))(5xx10^(-2))) = 9.33 xx 10^(-4) M`
`[CH_3COO^-] = [H^+] = 9.33 xx 10^(-4) M`
`pH = -log(9.33 xx 10^(-4)) = 4 - 0.9699 = 4 - 0.97 = 3.03`
Solution 2
Method 1
1) `CH_3COOH ↔ CH_3COO^(-) + H^(-) K_a = 1.74 xx 10^(-5)`
2) `H_2O + H_2O ↔ H_3O^(+) + OH^(-)` `K_w = 1.0 xx 10^(-14)`
Since Ka >> Kw, :
`CH_3COOH + H_2O ↔ CH_3COO^(-) +H_3O^(+)`
`C_i` = 0.05 0 0
`0.05 - 05alpha` `0.05alpha` `0.05alpha`
`K_a =((.05alpha) (.05 alpha))/((.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`
`0.05 alpha^2 + 1.74 xx 10^(-5) alpha - 1.74 xx 10^(-5)`
`D = b^2 - 4c`
`= (1.74 xx 10^(-5))^2 - 4(.05)(1.74 xx 10^(-5))`
`= 3.02 xx 10^(-23) + .348 xx 10^(-5)`
`alpha = sqrt(K_a)/c`
`alpha = sqrt(((1.74 xx 10^(-5)))/.05`
`= sqrt((34.8 xx 10^(-5) xx 10)/10)`
`= sqrt(3.48 xx 10^(-6))`
`= CH_3COOH ↔ CH_3COO^(-) + H^(+)`
= `(0.93xx 10^(-3))/1000`
= .000093
Method 2
Degree of dissociation,
`alpha = sqrt((K_a)/c)`
c = 0.05 M
`Ka = 1.74 xx 10^(-5)`
Then `alpha = sqrt((1.74 xx 10^(-5))/.05)`
`alpha = sqrt(34.8 xx 10^(-5))`
`alpha = 1.8610^(-2)`
`CH_3COOH ↔ CH_3COO^(-) + H^(+)`
Thus, concentration of CH3COO– = c.α
`= .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.
