Question

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.

Answer 1

1. \[\ce{CH_3COOH ↔ CH_3COO- + H-}\]   `"K"_"a" = 1.74 xx 10^(-5)`
2. \[\ce{H_2O + H_2O ↔ H_3O+ +  OH-}\] `"K"_"w" = 1.0 xx 10^(-14)`

Since Ka >> K_w, :

	CH3COOH	+	H2O	↔	CH3COO-	+	H3O+
Ci	0.05				0		0
	0.05 - 0.5 α				0.05 α		0.05 α

`"K"_"a" =((0.05  alpha) (0.05  alpha))/((0.05 - 0.05  alpha))`

`= ((0.05  alpha)(0.05  alpha))/(0.05(1-alpha))`

`= (0.05  alpha^2)/(1 - alpha)`

`1.74 xx 10^(-5) = (0.05  alpha^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)))/0.05`

`= sqrt((34.8 xx 10^(-5) xx 10)/10)`

`= sqrt(3.48 xx 10^(-6))`

\[\ce{CH_3COOH ↔ CH_3COO- + H+}\]

α1.86 × 10^-3

`["CH"_3"COO"^-] = 0.05 xx 1.86 xx 10^-3`  

= `(0.93xx 10^(-3))/1000`

= 0.000093

Answer 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))/0.05)`

`alpha = sqrt(34.8 xx 10^(-5))`

`alpha = 1.8610^(-2)`

\[\ce{CH_3COOH ↔ CH_3COO- + H+}\]

Thus, concentration of CH₃COO– = c.α

`= 0.05 xx 1.86 xx 10^(-2)`

`= 0.093 xx 10^(-2)`

= 0.00093  M

Since `["oAc"^(-)] = ["H"^+]`

`["H"^+] = 0.00093 = 0.093 xx 10^(-2)`

`"pH" = -log["H"^+]`

`= - log(0.093 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.