Question

How much of water must be added to 300 mL of 0.2 M solution of `"CH"_3"COOH"` for the degree of dissociation of the acid to be doubled? [`"K"_"a"` for acetic acid is `1.8 × 10^-5`]

1. 900 mL
2. 1000 mL
3. 500 mL
4. 300 mL

Answer 1

1. 900 ml

Explanation:

Ethanoic acid is a weak acid and dissociates:

`sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)`

For which `sf(K_a=([CH_3COO^(-)][H^(+)])/([CH_3COOH])=1.8xx10^(-5))`

These are equilibrium concentrations.

The degree of dissociation `sf(alpha)` is the fraction of the amount of electrolyte which dissociates. If C is the initial concentration of the acid then we can write an ICE table based on `sf("mol/l")`:

`sf(color(white)(xx)CH_3COOHcolor(white)(xxxx)rightleftharpoonscolor(white)(xx)CH_3COO^(-)color(white)(xxx)+color(white)(xxxx)H^(+))`

`sf(Icolor(white)(xxxxx)Ccolor(white)(xxxxxxxxxxxxxx)0color(white)(xxxxxxxxxxxxx)0)`

`sf(Ccolor(white)(xx)-Calphacolor(white)(xxxxxxxxxxx)+Calphacolor(white)(xxxxxxxxxxx)+Calpha)`

`sf(Ecolor(white)(xx)C-Calphacolor(white)(xxxxxxxxxxxx)Calphacolor(white)(xxxxxxxxxxxx)Calpha)`

`:.` `sf(K_a=(C^2alpha^2)/((C-Calpha))=(C^(2)alpha^2)/(C(1-alpha))=(Calpha^2)/((1-alpha))`

Because the dissociation is small i.e `sf(10^(-9)< K_a<10^(-4))` then we can assume that `sf((1-alpha)rArr1)`.

`:.` `sf(K_a=Calpha^2)`

`sf(alpha=sqrt(K_a/C))`

`sf(alpha=sqrt((1.8xx10^(-5))/(0.2))=9.486xx10^(-3))`

Now we double this value to get `sf(alpha^*)`:

`sf(alpha^*=2xx9.486xx10^(-3)=18.97xx10^(-3))`

To get the new concentration we get:

`sf(18.97xx10^(-3)=sqrt((1.8xx10^(-5))/C^*))`

`sf(360xx10^(-6)=(1.8xx10^(-5))/C^*)`

`sf(C^*=(1.8xx10^(-5))/(360xx10^(-6))=0.05color(white)(x)"mol/l")`

This is the new concentration we need. To find how much water we need to add to achieve this dilution we can say:

`sf(C_1V_1=C_2V_2)`

`:.` `sf(300xx0.2=0.05xxV_2)`

`sf(V_2=(300xx0.2)/(0.05)=1200color(white)(x)ml)`

This means we need to add a further 1200 - 300 = 900 ml. Which gives option 1.